The coefficient of determination
Multiple Choice
is the square root of the coefficient of correlation.
can range from -1.00 up to 1.00.
reports the proportion of variation in the dependent variable explained by changes in the independent variable.
measures the strength of the relationship between two variables.

Answers

Answer 1

Answer:

Step-by-step explanation:

is the square root of the coefficient of correlation.

can range from -1.00 up to 1.00.

reports the percent of the variation in the dependent variable explained by the independent variable.

is the strength of the relationship between two variables.


Related Questions

Find a value of the standard normal random variable z ​, call it z0​, such that the following probabilities are satisfied.
a.) P (z < z0​) = 0.0221
b.) P (-z0​ < z < z0​) = 0.95
c.) P (-z0​ < z < z0​) = 0.90
d.) P (-z0​ < z < z0​) = 0.8558
e.) P (-z0​ < z < 0)= 0.2949
f.) P (-2 < z < z0​) = 0.9503
g.) P (z > z0​) = 0.5
h.) P (z < z0​) = 0.0084

Answers

To find the value of the standard normal random variable z that satisfies the given probabilities, we can use a standard normal distribution table or a statistical calculator.

Since calculating all the probabilities manually would be time-consuming, I'll provide you with the values directly.

a.) P(z < z0) = 0.0221: The value of z0 is approximately -2.05.

b.) P(-z0 < z < z0) = 0.95: The value of z0 is approximately 1.96.

c.) P(-z0 < z < z0) = 0.90: The value of z0 is approximately 1.645.

d.) P(-z0 < z < z0) = 0.8558: The value of z0 is approximately 1.439.

e.) P(-z0 < z < 0) = 0.2949: The value of z0 is approximately -0.55.

f.) P(-2 < z < z0) = 0.9503: The value of z0 is approximately 1.96.

g.) P(z > z0) = 0.5: The value of z0 is approximately 0.

h.) P(z < z0) = 0.0084: The value of z0 is approximately -2.4.

Please note that the provided values are approximations based on common z-scores from the standard normal distribution table.

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P(z < z0) = 0.0084. Using the standard normal distribution table, we find that z0 is approximately -2.43.

To find the value of the standard normal random variable z (z0) that satisfies the given probabilities, we can use the standard normal distribution table or a calculator. Here are the solutions for each scenario:

a) P(z < z0) = 0.0221

Using the standard normal distribution table, we find that z0 is approximately -2.05.

b) P(-z0 < z < z0) = 0.95

Since the probability is symmetric, we can find the value of z0 by finding the z-value that corresponds to a cumulative probability of (1 - 0.95)/2 = 0.025. Using the standard normal distribution table, we find that z0 is approximately 1.96.

c) P(-z0 < z < z0) = 0.90

Similar to the previous scenario, we find the z-value that corresponds to a cumulative probability of (1 - 0.90)/2 = 0.05. Using the standard normal distribution table, we find that z0 is approximately 1.645.

d) P(-z0 < z < z0) = 0.8558

We find the z-value that corresponds to a cumulative probability of (1 - 0.8558)/2 = 0.0721. Using the standard normal distribution table, we find that z0 is approximately 1.43.

e) P(-z0 < z < 0) = 0.2949

We find the z-value that corresponds to a cumulative probability of 0.2949. Using the standard normal distribution table, we find that z0 is approximately -0.55.

f) P(-2 < z < z0) = 0.9503

We can subtract the cumulative probability of -2 from the desired probability Using the standard normal distribution table, we find that the cumulative probability for -2 is approximately 0.0228. Therefore, the remaining probability is 0.9503 - 0.0228 = 0.9275. We find the z-value that corresponds to this cumulative probability, which is approximately 1.75.

g) P(z > z0) = 0.5

Since the distribution is symmetric, the z-value that satisfies this probability is 0.

h) P(z < z0) = 0.0084

Using the standard normal distribution table, we find that z0 is approximately -2.43.

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4.000 high school freshmen from eight high schools took a common algebra exam. The scores were distributed normally with a mean of 75 and a standard deviation of 7. Use the T1-84, not the Empirical Rule, to answer this question. The number of students who had a score of at least 85 is Round answer to the nearest whole number. No comma. No space.

Answers

the number of students who had a score of at least 85 is approximately 1,368.

To find the number of students who had a score of at least 85, use the T1-84 calculator and the given information about the normal distribution.

Using the T1-84 calculator:

1. Press the "2nd" button, then "Vars" (DISTR).

2. Select "2: normalcdf(" from the options.

3. Enter the lower bound as 85, the upper bound as infinity (∞), the mean as 75, and the standard deviation as 7.

4. Press "Enter" to calculate the cumulative probability.

The result will give us the probability of a score being at least 85. To find the number of students, multiply this probability by the total number of students (4,000) and round the result to the nearest whole number.

Using the T1-84 calculator, the number of students who had a score of at least 85 is approximately 1,368.

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these selected numbers are the resul of the nomal selection procedure used for every drwing, the distrbution of the selected numbers is a normal dathbution? What is tho chape of the distribeiton of thoses welected numbers? Wil A be a normal detributco? Chacse the right arsiser. A. The distribubon will be rectangularshaped and nol a normat ditributon, 8. The disiributon we be cecular-shaped and not a normal distribution: C. The devibution wit be bell-shaped but not a normal distrbution. D. The datribution will be bel-shaped and n is a normal distributon

Answers

Option D is correct,  distribution will be bell-shaped and is not a normal distribution.

The distribution will be bell-shaped and is not a normal distribution.

The reason is that the selected numbers are the result of a normal selection procedure, which implies that they follow a normal distribution. The normal distribution is well-known for its bell-shaped curve.

Therefore, the distribution of the selected numbers will also be bell-shaped.

However, it's important to note that being bell-shaped does not necessarily mean that the distribution is a normal distribution.

The normal distribution has specific characteristics, such as a symmetric bell-shaped curve and specific mean and standard deviation values.

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Let f(x,y)={cx 5y 90 if 0≤x≤1,0≤y≤1otherwise Find the following: (a) c such that f(x,y) is a probability density function: c= (b) Expected values of X and Y : E(X)= E(Y)= (c) Are X and Y independent? (enter YES or NO)

Answers

The given function f(x, y) can be a probability density function if c is chosen as 36. The expected values of X and Y are both 0.5, indicating that they have equal averages. X and Y are not independent because the conditional distribution of Y depends on the value of X.

To find the value of c such that f(x, y) is a probability density function (PDF), we need to ensure that the integral of f(x, y) over its entire domain is equal to 1. In this case, the domain is the square region defined by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.

∫∫f(x, y)dxdy = ∫∫(cx + 5y + 90)dxdy

Evaluating the integral, we get:

∫∫(cx + 5y + 90)dxdy = c/2 + 5/2 + 90 = c/2 + 95/2

To satisfy the condition that the integral equals 1, we set c/2 + 95/2 = 1 and solve for c:

c/2 + 95/2 = 1

c/2 = 1 - 95/2

c/2 = 2/2 - 95/2

c/2 = -93/2

c = -93

Therefore, c = -93 for f(x, y) to be a probability density function.

To calculate the expected values of X and Y, we need to integrate x * f(x, y) and y * f(x, y) over their respective domains and then simplify the expressions:

b)E(X) = ∫∫x * f(x, y) dxdy

= ∫∫(c[tex]x^{2}[/tex] + 5xy + 90x) dxdy

= (c/3 + 45/2)

c)E(Y) = ∫∫y * f(x, y) dxdy

= ∫∫(cy + 5[tex]y^2[/tex] + 90y) dxdy

= (c/2 + 95/3)

Therefore, E(X) = (c/3 + 45/2) and E(Y) = (c/2 + 95/3).

To determine if X and Y are independent, we need to check if the joint PDF can be factored into the product of the marginal PDFs of X and Y. In this case, it is clear that f(x, y) cannot be separated into independent functions of x and y. Therefore, X and Y are not independent.

Overall, (a) c = -93, (b) E(X) = (c/3 + 45/2) and E(Y) = (c/2 + 95/3), and (c) X and Y are not independent.

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In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 11 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x 8.11 cm and -0.8 cm. Conduct the appropriate test at the a0.01 level of significance.
State the null and alternative hypotheses
H
(Type integers or decimals. Do not round)
Identify the 1-statistic
(Round to two decimal places as needed)
Identify the P-value.
(Round to three decimal places as needed)
Make a conclusion regarding the hypothesis
the null hypothesis. There
sufficient evidence to claim that the mean hippocampal volume is

Answers

Based on the provided options, the conclusion would be:

Reject the null hypothesis. There is sufficient evidence to claim that the mean hippocampal volume is less than 9.02 cm³.

The null and alternative hypotheses for this study are:

Null Hypothesis (H0): The mean hippocampal volume in adolescents with alcohol use disorders is equal to 9.02 cm³.

Alternative Hypothesis (H1): The mean hippocampal volume in adolescents with alcohol use disorders is less than 9.02 cm³.

To conduct the appropriate test at the 0.01 level of significance, we will perform a one-sample t-test.

To calculate the t-statistic, we can use the formula:

t = (X- μ) / (s / √n)

where X is the sample mean, μ is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size.

Given the sample data:

X = 8.08 cm³

μ = 9.02 cm³

s = 0.7 cm³

n = 10

Substituting the values into the formula:

t = (8.08 - 9.02) / (0.7 / √10) ≈ (-5.78)

The t-statistic is approximately (-5.78).

To determine the p-value, we need to consult the t-distribution table or use software/calculator. Based on the t-statistic and the degrees of freedom (n - 1 = 10 - 1 = 9), the p-value can be obtained.

However, if the p-value is less than 0.01 (the significance level), we would reject the null hypothesis. If the p-value is greater than or equal to 0.01, we would fail to reject the null hypothesis.

Based on the provided options, the conclusion would be:

Reject the null hypothesis. There is sufficient evidence to claim that the mean hippocampal volume is less than 9.02 cm³.

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--The question is incomplete, the given complete question is:

"In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 10 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm cubed. An analysis of the sample data revealed that the hippocampal volume is approximately normal with x =8.08 cm cubed and s=0.7 cm cubed. Conduct the appropriate test at the 0.01 level of significance. State the null and alternative hypotheses.

Upper H 0​: mu equals 9.02 Upper H 1​: mu less than 9.02

Identify the​ t-statistic. ​(Round to two decimal places as​ needed.

Identify the​ P-value. ​P-value​(Round to three decimal places as​ needed

Make a conclusion regarding the hypothesis.  Fail to reject. Reject the null hypothesis. There is not sufficient evidence to claim that the mean hippocampal volume is equal to less than greater than nothing cm cubed."--

Given the LTI (linear-time-invariant) system with the "triangular" impulse response h(t)= (1-1-41) 0 st≤ 24 t<0 and t> 2A Calculate the Fourier integral H (w) of h(t) and draw h(t) and the absolute value of H (w) schematically.

Answers

To calculate the Fourier integral H(w) of the impulse response h(t), we can use the definition of the Fourier transform.

The Fourier transform is defined as:

H(w) = ∫[h(t) * e^(-jwt)] dt

First, let's consider the interval t ≤ 0:

For t ≤ 0, h(t) = 0, so the integral becomes:

H(w) = ∫[0 * e^(-jwt)] dt = 0

Next, let's consider the interval 0 < t ≤ 2:

For 0 < t ≤ 2, h(t) = 1, so the integral becomes:

H(w) = ∫[1 * e^(-jwt)] dt = ∫e^(-jwt) dt

Integrating e^(-jwt) with respect to t gives:

H(w) = [-j/w * e^(-jwt)] | from 0 to 2

Plugging in the limits of integration, we have:

H(w) = [-j/w * e^(-2jw) + j/w * e^(0)] = -j/w * (e^(-2jw) - 1)

Finally, let's consider the interval t > 2:

For t > 2, h(t) = 0, so the integral becomes:

H(w) = ∫[0 * e^(-jwt)] dt = 0

Therefore, the Fourier integral H(w) is:

H(w) = -j/w * (e^(-2jw) - 1) for 0 < w ≤ 2

H(w) = 0 for w > 2 and w ≤ 0

To draw the schematic representation, we can plot the impulse response h(t) and the absolute value of H(w) on separate graphs.

For h(t):

The impulse response h(t) is 0 for t ≤ 0 and t > 2.

From 0 < t ≤ 2, h(t) is a triangle shape with a height of 1.

Draw a straight line connecting (0, 0) to (2, 1) and continue the line as 0 for t > 2 and t ≤ 0.

For |H(w)|:

The absolute value of H(w) is 0 for w > 2 and w ≤ 0.

For 0 < w ≤ 2, |H(w)| is a constant value of |H(w)| = |(-j/w * (e^(-2jw) - 1))|.

Mark the height of |H(w)| for 0 < w ≤ 2.

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Question 7
I asked in the week one questionnaire how many of you agreed with the following statement "Activities of married women are best confined to the home and family". I received 376 responses to the survey.This is an example of a measure of social conservatism. My research question is: Is there is a difference in the levels of social conservatism between men and women?
Agree Disagree
Female 31 269
Male 7 66
Prefer not to say 0 3
What is the conditional probability that if it is a male that they agree with the statement to 3 significant figures?
1. 0.01892
2. 0.0189
3. 0.0959
4.0.096

Answers

The conditional probability that a male agrees with the statement is,

⇒ 0.0959.

Hence option 3 is correct.

To find the conditional probability that a male agrees with the statement, we have to divide the number of male respondents who agree with the statement by the total number of male respondents who answered the question.

From the data you provided, we have,

Number of male respondents who agree with the statement = 7

Total number of male respondents who answered the question = 7+66

                                                                                                           = 73

Therefore, the conditional probability that a male agrees with the statement is,

P(Agree|Male) = number of male respondents who agree with the statement / total number of male respondents who answered the question

= 7 / 73

= 0.0959

Hence, the correct option would be 0.0959.

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Let z = z(x, y) be an implicit function defined by the equation x^3 + 3(y^2)z − xyz^3 = 0. Find ∂z/∂x and ∂z/∂y .

Answers

The partial derivatives ∂z/∂x and ∂z/∂y of the implicit function z = z(x, y) defined by the equation x^3 + 3(y^2)z − xyz^3 = 0 are given by ∂z/∂x = (yz^3 - 3x^2) / (3(y^2) - 3xz^2) and ∂z/∂y = (xz^3 - 6yz) / (3(y^2) - 3xz^2), respectively.

To find the partial derivative ∂z/∂x, we differentiate the equation x^3 + 3(y^2)z − xyz^3 = 0 with respect to x, treating z as a function of x and y. Rearranging the terms and solving for (∂z/∂x), we obtain ∂z/∂x = (yz^3 - 3x^2) / (3(y^2) - 3xz^2).

Similarly, to find the partial derivative ∂z/∂y, we differentiate the equation with respect to y, treating z as a function of x and y. Rearranging the terms and solving for (∂z/∂y), we obtain ∂z/∂y = (xz^3 - 6yz) / (3(y^2) - 3xz^2).

Therefore, the partial derivatives are ∂z/∂x = (yz^3 - 3x^2) / (3(y^2) - 3xz^2) and ∂z/∂y = (xz^3 - 6yz) / (3(y^2) - 3xz^2).

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QUESTION 5
It s knovm that mean age of me population ot n same community 51 years. What s the expected valuen of the sampling distibutm of the sanple mean? Round your answer to the nunber of years
QUESTION 6
If me proporton of Mutercard tansactms in a of various credit card transactions iS ecuaj to 0.276 and samoe size iS n. wtat iS the margin ot error for the corresoonding 94% wmnoence Reval to estmate the
propotion of Mastercard transactons in the population? Assume that me conditions for the sampling distribution to be approximately normal are satisfied. Use Excel to calculate and round your answer to 4 decimals.

Answers

5.  The mean age of the population of doctors is known to be 53 years, so the expected value of the sampling distribution of the sample mean is also 53 years.

6. The left boundary of the confidence interval is 57.13.

5. To find the left boundary of the confidence interval to estimate the population mean, we need to subtract the margin of error from the sample mean.

Left boundary = Sample mean - Margin of error

Given:

Sample mean = 62.66

Margin of error = 5.53

6. Calculating the left boundary:

Left boundary = 62.66 - 5.53 = 57.13

Therefore, the left boundary of the confidence interval to estimate the population mean is 57.13.

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The complete question is below:

It s known that the mean age of the populaiton of doctors in some community is 53 years. What is the expected value of the sampling distribution of the sample mean? Round your answer to the whole number of years. QUESTION 6 Given a sample mean 62.66 and a margin of error 5.53, what is the left boundary of the confidence interval to estimate the population mean? Round your answer to 2 decimal places.

A coin-operated drink machine was designed to discharge a mean of 6 fuld ounces of coffee per cup. In a test of the machine, the charge atsi31 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.13 fuid ounces and 0.31 ounces, respectively If we assume that the discharge amounts are approximately normally distributed, is there enough evidence, to conclude that the pripulation mean discharge, differs from 6 fluid ounces? Use the 0.05 level of significance. Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consulta list of formulas) (a) State the null hypothesis , and the alternative hypothesis 10 (b) Determine the type of test statistic to use.

Answers

In this hypothesis test, the null hypothesis states that the population mean discharge is equal to 6 fluid ounces, while the alternative hypothesis suggests that the population mean discharge differs from 6 fluid ounces.

To test this, we will use a two-tailed test at a significance level of 0.05.

(a) The null hypothesis (H0) and the alternative hypothesis (Ha) can be stated as follows:

Null hypothesis (H0): The population mean discharge is equal to 6 fluid ounces.

Alternative hypothesis (Ha): The population mean discharge differs from 6 fluid ounces.

(b) To test these hypotheses, we will use a two-tailed test because the alternative hypothesis does not specify whether the population mean discharge is greater or smaller than 6 fluid ounces. We want to determine if there is evidence to conclude that the population mean discharge is different from 6 fluid ounces.

To perform the hypothesis test, we need to calculate the test statistic. In this case, since the sample size is large (n > 30) and the population standard deviation is unknown, we will use the t-test statistic. The formula for the t-test statistic is:

t = (sample mean - population mean) / (sample standard deviation / √n)

where t follows a t-distribution with (n - 1) degrees of freedom. We will compare the calculated t-value with the critical t-value from the t-distribution table, considering a two-tailed test at a significance level of 0.05. If the calculated t-value falls outside the critical region, we can reject the null hypothesis and conclude that the population mean discharge differs from 6 fluid ounces.

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the question is " find the radious and interval of convergence of
the following series,
can you make this question on paper and step by step please ?

Answers

The radius and interval of convergence of the given series [tex]\sum_{k=1}^\infty[/tex] (x - 2)ᵏ . 4ᵏ are 0.25 and (1.75, 2.25) respectively.

Given the series is (x - 2)ᵏ . 4ᵏ

So the k th term is = aₖ = (x - 2)ᵏ . 4ᵏ

The k th term is = aₖ₊₁ = (x - 2)ᵏ⁺¹ . 4ᵏ⁺¹

So now, | aₖ₊₁/aₖ | = | [(x - 2)ᵏ⁺¹ . 4ᵏ⁺¹]/[(x - 2)ᵏ . 4ᵏ] | = | 4 (x - 2) |

Since the series is convergent then,

| aₖ₊₁/aₖ | < 1

| 4 (x - 2) | < 1

- 1 < 4 (x - 2) < 1

- 1/4 < x - 2 < 1/4

- 0.25 < x - 2 < 0.25

2 - 0.25 < x - 2 + 2 < 2 + 0.25 [Adding 2 with all sides]

1.75 < x < 2.25

So, the radius of convergence = 1/4 = 0.25

and the interval of convergence is (1.75, 2.25).

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Complete question is below

find the radius and interval of convergence of the following series

[tex]\sum_{k=1}^\infty[/tex] (x - 2)ᵏ . 4ᵏ

5. The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 0.125 inches. The width of a door is normally distributed with a mean of 23.875 inches and a standard deviation of 0.0625 inches. Assume independence. (10) (a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door? (10) (b) What is the probability that the width of the casing minus the width of the door exceeds 0.25 inches?

Answers

(a) The mean of the difference between the width of the casing and the width of the door is 0.125 inches, and the standard deviation is  0.1397 inches.

(b) The probability that the width of the casing minus the width of the door exceeds 0.25 inches is 0.1841 or 18.41%.

(a) Given the mean of casing width (X₁) = 24 inches

Standard deviation of casing width (σ₁) = 0.125 inches

Mean of door width (X₂) = 23.875 inches

Standard deviation of door width (σ₂) = 0.0625 inches

The difference between the width of the casing and the width of the door can be represented as:

Difference (X₁- X₂) = X₁- X₂

The mean of the difference is equal to the difference in means:

Mean of difference = Mean(X₁- X₂)

= Mean(X₁) - Mean(X₂) = 24 - 23.875

= 0.125 inches.

The variance of the difference is the sum of the variances:

Variance of difference = Variance(X₁) + Variance(X₂)

= (σ₁²) + (σ₂²) = (0.125²) + (0.0625²)

= 0.015625 + 0.00390625

= 0.01953125

The standard deviation of the difference is the square root of the variance:

Standard deviation of difference = √(Variance of difference) = √(0.01953125)

= 0.1397 inches.

(b) To find the probability that the width of the casing minus the width of the door exceeds 0.25 inches.

we need to calculate the z-score and then find the corresponding probability from the standard normal distribution.

Let us find the z-score:

z = (x - μ) / σ

x = 0.25 inches, μ = 0.125 inches, and σ = 0.1397 inches.

z = (0.25 - 0.125) / 0.1397

= 0.895

Now, we need to find the probability corresponding to a z-score of 0.895.

Using a standard normal distribution table , we can find this probability. Let's assume it is denoted by P(Z > 0.895).

P(Z > 0.895) = 1 - P(Z < 0.895)

Using the standard normal distribution table , we find that P(Z < 0.895) = 0.8159.

Therefore, P(Z > 0.895) = 1 - 0.8159

= 0.1841.

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Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 381 with 74% successes at a confidence level of 99.8%. M.E. =% Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Round final answer to one decimal place

Answers

The margin of error (M.E.) corresponding to a sample of size 381 with 74% successes at a confidence level of 99.8% is approximately 3.5%.

To find the margin of error (M.E.), we need to consider the sample size, the proportion of successes in the sample, and the confidence level.

Calculate the critical value (z-score) for a 99.8% confidence level:

The confidence level of 99.8% corresponds to a significance level of (1 - 0.998) = 0.002. Since the confidence level is high, we can assume a normal distribution. Looking up the critical value for a two-tailed test with a significance level of 0.002 in the standard normal distribution table, we find a value of approximately 3.09 (rounded to 3 decimal places).

Calculate the standard error (SE):

The standard error measures the variability of sample proportions around the true population proportion. It can be calculated using the formula: SE = sqrt((p * (1 - p)) / n), where p is the sample proportion and n is the sample size. Substituting the values, we have: SE = sqrt((0.74 * 0.26) / 381) ≈ 0.026.

Calculate the margin of error (M.E.):

The margin of error represents the maximum likely difference between the sample proportion and the true population proportion. It can be calculated by multiplying the critical value (z-score) by the standard error. Thus, M.E. = z * SE ≈ 3.09 * 0.026 ≈ 0.08034. Rounded to one decimal place, the margin of error is approximately 0.1 or 3.5%.

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Solve the system: 3x−5y=a
5x−7y=b
​where (i) a=0,b=0 and (ii) a=2,b=6 What is the solution of only the first equation in (i) and (ii), and what is the solution of the second equation in (i) and (ii)?

Answers

Given system of equations are:

3x-5y = a  --- (1)  

5x - 7y = b  --- (2)

(i) a = 0, b = 0

Substituting the given values in the above equations, we get

3x - 5y = 0 --- (1)

5x - 7y = 0  --- (2)

Now, let's solve the equations to get the solutions:

For equation 1:

3x - 5y = 0  

⇒ 3x = 5y

keeping x on the other side by dividing the entire equation by 3

⇒ x = (5/3)y

So, the solution of the first equation is x = (5/3)y

For equation 2:

5x - 7y = 0

⇒ 5x = 7y  

keeping x on the other side by dividing the entire equation by 5

⇒ x = (7/5)y  

So, the solution of the second equation is x = (7/5)y

(ii) a = 2, b = 6

Substituting the given values in the above equations, we get

3x - 5y = 2  --- (1)  

5x - 7y = 6  --- (2)  

Now, let's solve the equations to get the solutions:

For equation 1:

3x - 5y = 2  

⇒ 3x = 5y + 2  

keeping x on the other side by dividing the entire equation with 3

⇒ x = (5/3)y + (2/3)  

So, the solution of the first equation is x = (5/3)y + (2/3)

For equation 2:

5x - 7y = 6  

⇒ 5x = 7y + 6  

keeping x on the other side by dividing the entire equation by 5

⇒ x = (7/5)y + (6/5)  

So, the solution of the second equation is x = (7/5)y + (6/5)

Hence, the solutions of the first equation in (i) and (ii) are: x = (5/3)y and x = (5/3)y + (2/3) respectively.

And the solutions of the second equation in (i) and (ii) are: x = (7/5)y and x = (7/5)y + (6/5) respectively.

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Lucy is a dress maker. She dews 4/7 of a dress in 0.75 hours. At
this rate, how many dresses does Lucy sew in one hour? (Include
fractions of dresses if applicable)

Answers

Lucy sews 16/21 of a dress in one hour. also included fractions .

To find out how many dresses Lucy sews in one hour, follow these steps:

1. Start with the given information: Lucy sews 4/7 of a dress in 0.75 hours.

2. To determine the rate of sewing per hour, we need to find the ratio of dresses sewn to the time taken.

3. Divide the fraction of a dress (4/7) by the time in hours (0.75): (4/7) / 0.75.

4. To divide fractions, multiply the first fraction by the reciprocal of the second fraction: (4/7) * (1/0.75).

5. Simplify the fraction multiplication: (4/7) * (4/3) = 16/21.

6. The simplified fraction 16/21 represents the number of dresses Lucy sews in one hour.

7. Therefore, Lucy sews 16/21 of a dress in one hour.

8. If needed, this can also be expressed as a mixed number or decimal for practical purposes.

In summary, Lucy sews 16/21 of a dress in one hour based on the given information.

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Oxnard Petro, Ltd., has three interdisciplinary project development teams that function on an ongoing basis. Team members rotate from time to time. Every 4 months (three times a year) each department head rates the performance of each project team (using a 0 to 100 scale, where 100 is the best rating). Are the main effects significant? Is there an interaction?
Year Department
Marketing Engineering Finance
2004 90 69 96
84 72 86
80 78 86
2005 72 73 89
83 77 87
82 81 93
2006 92 84 91
87 75 85
87 80 78
Choose the correct row-effect hypotheses.
a. H0: A1 ≠ A2 ≠ A3 ≠ 0 H1: All the Aj are equal to zero
b. H0: A1 = A2 = A3 = 0 H1: Not all the Aj are equal to zero
(a-2) Choose the correct column-effect hypotheses.
a. H0: B1 ≠ B2 ≠ B3 ≠ 0 H1: All the Bj are equal to zero
b. H0: B1 = B2 = B3 = 0 H1: Not all the Bj are equal to zero
(a-3) Choose the correct interaction-effect hypotheses.
a. H0: Not all the ABjk are equal to zero H1: All the ABjk are equal to zero
b. H0: All the ABjk are equal to zero H1: Not all the ABjk are equal to zero

Answers

The row-effect hypotheses compare department means, the column-effect hypotheses compare year means, and the interaction-effect hypotheses examine interaction effects.



To determine the main effects and interaction in the given data, we can perform a two-way analysis of variance (ANOVA). The row effect corresponds to the three departments (Marketing, Engineering, Finance), the column effect corresponds to the three years (2004, 2005, 2006), and the interaction effect tests whether the combined effect of department and year is significant.The correct row-effect hypotheses are:

a- H0: A1 ≠ A2 ≠ A3 ≠ 0 (Null hypothesis: the means of the departments are not all equal)b- H1: All the Aj are equal to zero (Alternative hypothesis: the means of the departments are all equal)

The correct column-effect hypotheses are:b- H0: B1 = B2 = B3 = 0 (Null hypothesis: the means of the years are all equal)

a- H1: Not all the Bj are equal to zero (Alternative hypothesis: the means of the years are not all equal)The correct interaction-effect hypotheses are:

b- H0: All the ABjk are equal to zero (Null hypothesis: there is no interaction effect)a- H1: Not all the ABjk are equal to zero (Alternative hypothesis: there is an interaction effect)

To determine if the main effects and interaction are significant, we would need to perform the ANOVA calculations using statistical software or tables and compare the obtained p-values with a chosen significance level (e.g., α = 0.05).

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If we like to test
H0: Variance of electronic = Variance of utilities
H1: Variance of electronic ≠ Variance of utilities
Which one is the correct one?
a. We need to do Shapiro test with log transformed data.
b. Since we do not assume that both data are from a normal distribution, we need to do Shapiro test first to see that data are from a normal distribution.
c. We can do F test right away.
d. We need to do F test with log transform data.

Answers

If the F statistic is less than the critical value, we fail to reject the null hypothesis and conclude that the variances are equal.So, the correct answer is c.

An F-test is a statistical test used to compare the variances of two or more samples.

It is used to compare whether the variances of two groups are similar or different, or whether the variances of multiple groups are equal or different.

In ANOVA, the F-test is used to determine whether there is a significant difference between the means of two or more groups. It is also used to test for the significance of regression models.In this case, we need to test the following hypotheses:H0:

Variance of electronic = Variance of utilitiesH1:

Variance of electronic ≠ Variance of utilitiesFor this case, we can perform an F-test right away to determine whether the variances are equal or not.

The F-test for equality of variances is a one-tailed test.

It is calculated as the ratio of the variances of two samples.

The F statistic is calculated by dividing the variance of the sample with the larger variance by the variance of the sample with the smaller variance.

If the F statistic is greater than the critical value, we can reject the null hypothesis and conclude that the variances are not equal.

If the F statistic is less than the critical value, we fail to reject the null hypothesis and conclude that the variances are equal.So, the correct answer is c. We can do F test right away.

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3 12. X'= 2 x²=(²³₁ -¯3) x + (1²) X -1 con X(0) = -0

Answers

The given problem involves a first-order nonlinear ordinary differential equation (ODE). We are asked to solve the ODE with an initial condition. The equation is represented as X' = 2x² - (2³₁ - ¯3)x + (1²)x - 1, with the initial condition X(0) = -0.

To solve the given ODE, we can rewrite it as X' = 2x² - (8 - 3)x + (1)x - 1. Simplifying further, we have X' = 2x² - 5x + 1 - 1. This reduces to X' = 2x² - 5x.

To find the solution, we can proceed by separating variables and integrating both sides of the equation. Integrating the left side gives us ∫dX = ∫2x² - 5x dx. Integrating the right side yields X = (2/3)x³ - (5/2)x² + C, where C is the constant of integration.

Applying the initial condition X(0) = -0, we can substitute x = 0 into the equation and solve for C. Since the initial condition implies X(0) = 0, we find C = 0.

Therefore, the solution to the ODE is X = (2/3)x³ - (5/2)x².

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Translating research questions into hypotheses. Translate each of the following research questions into appropriate H_{0} and H_{a}
a. U.S. Census Bureau data show that the mean household income in the area served by a shopping mall is $78,800 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population.
b. Last year, your online registration technicians took an average of 0.4 hour to respond to trouble calls from students trying to register. Do this year's data show a different average response time?
c. In 2019, Netflix's vice president of original content stated that the average Netflix subscriber spends two hours a day on the service.15 Because of an increase in competing services, you believe this average has declined this year.

Answers

In order to conduct effective research, it is important to translate research questions into appropriate hypotheses. Hypotheses provide a clear and testable statement of what the researcher believes to be true about the population being studied.

For the first research question, the null hypothesis states that the mean household income of mall shoppers is not higher than that of the general population, while the alternative hypothesis suggests that the mean household income of mall shoppers is higher than that of the general population.

This allows the market research firm to investigate whether mall shoppers have a higher income level than the general population.

The second research question compares last year's average response time for online registration technicians with this year's data. The null hypothesis states that the average response time this year is the same as last year, while the alternative hypothesis suggests that the average response time this year is different from last year.

This enables the organization to determine if there has been any change in the performance of their technicians.

Finally, the third research question investigates if Netflix subscribers' average time spent on the service per day has declined this year. The null hypothesis states that the average time spent has not declined, while the alternative hypothesis suggests that it has.

This provides insight into how competing services may be impacting Netflix's user engagement.

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a) Let X be the random variable representing the number of accidents in a certain intersection in a week. The probability distribution of X is shown in TABLE 2. TABLE 2 Referring to TABLE 2, i. find the value of M. ii. present the information in TABLE 2 into a probability distribution graph. iii. find P(4

Answers

The probability of having exactly 4 accidents in a certain intersection in a week is 1/32 or 0.03125.

a) Let X be the random variable representing the number of accidents in a certain intersection in a week.

The probability distribution of X is shown in TABLE 2.

TABLE 2ValueProbability1 3/82 2/83 1/164 1/32i. find the value of M.The sum of the probabilities must be equal to 1. Hence,

we need to add all probabilities in the table 2 to find the value of M. We have:

ValueProbability13/82/83/161/32Adding all these probabilities, we get:M = 3/8 + 2/8 + 1/16 + 1/32= 6/16 + 4/16 + 1/16 + 1/32= 11/32ii. present the information in TABLE 2 into a probability distribution graph.

We will construct a probability distribution graph using the values from table 2,

as shown below:iii. find P(4)To find P(4), we need to look at the table 2.

We have:P(4) = 1/32Therefore,

the probability of having exactly 4 accidents in a certain intersection in a week is 1/32 or 0.03125.

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Based on the figure of post glacial sea level rise since the most recent retreat of the North American ice sheet, approximately how much lower was global sea level about 10000 years ago when the Holocene began?

Answers

The figure of post-glacial sea level rise since the most recent retreat of the North American ice sheet shows that global sea levels were significantly lower about 10,000 years ago when the Holocene began.

During this time, much of the Earth's water was still locked in ice sheets and glaciers from the previous ice age.

Estimates suggest that global sea levels were around 120 meters (394 feet) lower than present-day levels during the peak of the last ice age, which occurred around 20,000 years ago. By the time the Holocene began approximately 10,000 years ago, significant melting had already occurred, resulting in a rise in sea levels.

While the exact amount by which global sea levels were lower 10,000 years ago may vary depending on specific regional factors, it is generally estimated that sea levels were still considerably lower than today, likely by several tens of meters.

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In a shop near a school, pencils and erasers are sold. Pens are sold in packs of 10 and erasers in packs of 12. A student decides to buy the minimum number of packages of each variety that results in as many pens as erasers. How many packages should the student buy each? Use correct mathematical language.

Answers

The student should buy 6 packages of pens and 5 packages of erasers to get equal numbers of pens and erasers.

A student decides to buy the minimum number of packages of each variety, which results in as many pens as erasers. Pens are sold in packs of 10, and erasers in packs of 12. To determine the minimum number of packages of each type that will result in equal numbers of pens and erasers, the smallest common multiple of 10 and 12 must be calculated.

We can begin the process of finding the smallest common multiple of 10 and 12 by writing down their multiples:

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120 ...

Multiples of 12:

12, 24, 36, 48, 60, 72, 84, 96, 108, 120 ...

The smallest number that appears in both lists is 60. Therefore, 60 is the smallest common multiple of 10 and 12.

Since pens come in packages of 10 and erasers come in packages of 12, we need to find how many packages are required to obtain 60 of each. If we divide 60 by 10, we get 6.

This means we need to buy 6 packages of pens to get 60. If we divide 60 by 12, we get 5. This means we need to buy 5 packages of erasers to get 60. Hence the student should buy 6 packages of pens and 5 packages of erasers.

Therefore, the student should buy 6 packages of pens and 5 packages of erasers to get equal numbers of pens and erasers.

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points wm a 96\% level of confidence. Corplete parts (a) through (c) below. a. State the-survey resuts in confidence interval form and interofet the interval The confidence interval of the survey tecults is (Round to two decimai places as needed) Inteepret the intenval. Choose the cotrect answer below. A. There is a 60% chance that the percentage of people in the region who have tried marjuana is withir the confidence interval B. Wo are 80% confdent that the percentage of people in the region who have tried marijuana is within the confidence interval c. The confidence interval will cortain the percertage of people in the regon who have tred marijuana 90% of the time. D. 90% of the 1,040 people from the region that were polled fell within the confidence interval. b. If the poling company was to conduct 200 such surveys of 1,040 people tom the tegion, how many of them would rosult in confidence intervals that did not include the true population proporton? Approximately of the conlidence intervals woud not include the true populaton proportion anything, is incorrect in this imerpretation? A. This irterpretasion is incotrect becalase the confidence lovel states the probabilify that the sample proportion is within the confidence inienval. B. This interpretation is inconect because a confidenco intneval is about a populaton not a sanple. C. This interpretation is incorrect because the confidence level represents how often the confdence interval wil contain the coirect population proporion. D. There is nothing wong with this interpectation.

Answers

a. The confidence interval of the survey results is (58.67%, 61.33%).

b. 8 surveys would result in confidence intervals that did not include the true population proportion.

c. This interpretation is incorrect.

a. The confidence interval of the survey results is (58.67%, 61.33%).

Interpretation: We are 96% confident that the true percentage of people in the region who have tried marijuana falls within the range of 58.67% to 61.33%.

b. If the polling company conducted 200 surveys of 1,040 people from the region, 8 surveys would result in confidence intervals that did not include the true population proportion.

c. This interpretation is incorrect because the confidence level represents the probability that the confidence interval will contain the true population proportion, not the percentage of times it will contain it.

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Tina spent 3/5 of her money to buy a new pair of running shoes. If she paid $21 for the shoes, how much money did she have in the beginning? Write a multiplication equation involving fraction to represent the situation then solve it. Use any variable to represent the unknown value.

Answers

Tina initially had $35. She spent 3/5 of her money to buy the running shoes, which amounted to $21.

Let's assume Tina's initial amount of money is represented by the variable "M".

According to the given information, Tina spent 3/5 of her money to buy the running shoes, which is equivalent to paying $21. We can write this as a multiplication equation involving fractions:

(3/5) * M = 21

To solve this equation, we can isolate the variable M by dividing both sides of the equation by (3/5):

M = 21 / (3/5)

To divide by a fraction, we can multiply by its reciprocal:

M = 21 * (5/3)

Simplifying:

M = 35

Therefore, Tina had $35 in the beginning.

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Biologists are treating a lake contaminated with bacteria. The level of contamination is changing at a rate of day, where is the number of days since treatment began. Find a function N() to estimate the level of contamination if the level after 1 day was 5730 bacteria per cubic centimeter.

Answers

The level of contamination after 1 day was 5730 bacteria per cubic centimeter, the function N(t) can be written as N(t) = N₀ * e^(-kt), where N₀ represents the initial level of contamination and k is the decay constant.

To estimate the level of contamination in the lake, an exponential decay model is commonly used. In this case, the function N(t) represents the level of contamination at time t, and it can be expressed as N(t) = N₀ * e^(-kt). The value of N₀ is the initial level of contamination, and k represents the decay constant.

Given that the level of contamination after 1 day was 5730 bacteria per cubic centimeter, we can substitute the values into the exponential decay model equation:

5730 = N₀ * e^(-k * 1).

To determine the function N(t) and estimate the level of contamination at any given time, we would need more information, such as the decay rate of the contamination or additional data points to solve for the values of N₀ and k.

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The Wellington company wants to develop a simple linear regression model for one of its products. Use the following 12 periods of historical data to develop the regression equation and use it to forecast the next three periods.
The simple linear regression line is ??????????? ​(Enter your responses rounded to two decimal places and include a minus sign if​ necessary.)
Find the forecasts for periods​ 13-15 based on the simple linear regression and fill in the table below ​(enter your responses rounded to two decimal​ places).
Period (x) Forecast (Ft)
13 14 15

Answers

The simple linear regression line is 973.65 + ( -45.16 )[tex]x_{1}[/tex]

Forecast 13 = 386.57

Forecast 14 =  341.41

Forecast 15 = 296.25

Given,

12 periods of historical data.

Now,

According to simple regression line standard form,

y = mx + b

y =  response (dependent) variable

x = predictor (independent) variable

m = estimated slope

b = estimated intercept.

So here the the regression line equation will be

973.65 + (-45.16)[tex]x_{1}[/tex]

Forecast 13

Substitute the value of  [tex]x_{1}[/tex]

Forecast 13  = 973.65 + (-45.16)13

Forecast 13 = 386.57

Forecast 14

Substitute the value of  [tex]x_{1}[/tex]

Forecast 14  = 973.65 + (-45.16)14

Forecast 14 = 341.41

Forecast 15

Substitute the value of  [tex]x_{1}[/tex]

Forecast 15  = 973.65 + (-45.16)15

Forecast 15 = 296.25

Thus the regression line equations and forecast can be calculated .

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Suppose that the mean weight of newborn babies is normally distributed with a mean of 7.4 pounds and a standard deviation of 0.6 pound. A developmental psychologist wants to test whether newborn babies of mothers who use drugs during pregnancy differ in weight from the average baby. The psychologist takes a random sample of 30 mothers who used drugs during pregnancy and computes the mean birth weight of these mothers’ babies. This sample of 30 mothers has a sample mean of 7.0 pounds. Using an alpha level of .01, test whether mothers’ drug use during pregnancy affects newborn babies’ birth weight.
What is the standard error of the sampling distribution for this test?
Answer format: Number: Round to: 2 decimal places.

Answers

There is enough evidence to support the claim that newborn babies of mothers that consume drugs during pregnancy differs from average weight of new born .

Given,

That the mean weight of newborn babies is normally distributed with a mean of µ = 7.4 pounds and a standard deviation  = 0.7 pounds.

a) Mean for sampling distribution = 6.6 pounds.

b) Research for the average weight of new born babies.  Thus the hypothesis are:

[tex]H_{0}[/tex]: µ = 7.4

[tex]H_{a}[/tex]: µ ≠ 7.4

Depending on the hypothesis it will be a two tailed test .

Given a sample of n = 30 mothers has a sample mean of X  = 6.6 pounds.

Since the sample size is equal to 30 and the population standard deviation is known hence Z statistic is applicable for hypothesis testing.

Rejection region:

Reject [tex]H_{0}[/tex] if P-value is less than 0.01.

Test statistic:

Z = X - µ/σ/[tex]\sqrt{n}[/tex]

Z = 6.6 - 7.4 /0.7/[tex]\sqrt{30}[/tex]

Z = -6.26

P-value:

The P-value is computed using the excel formula for normal distribution , thus the p-value is computed as less than 0.01.

Since

The P-value is less than 0.01 hence we can reject the null hypothesis and conclude that there is enough evidence to support the claim that newborn babies of mothers that consume during pregnancy differs from average weight of new born .

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what is the correlation

Answers

Answer: moderate, C, B

Step-by-step explanation:

A)  A shows a moderate negative correlation.  It is moderate because the scattered points are sort of close to the line so it has moderate/medium correlation.  It is also negative because it has a negative slope

B) C shows the strongest correlation because the points around the line are tight and close.

C)  B should not have been drawn.  The  correlation is very weak.  You do know where the line should be because the points are all over the place.

A random sample of 40 observations is to be drawn from a large population of measurements. It is known that 30% of the measurements in the population are 1's, 20% are 2's, 20% are 3's and 30% are 4's.
a) Give the mean and standard deviation of the (repeated) sampling distribution of x, the sample mean of the 40 observations.
b) Describe the shape of the sampling distribution of . Does the answer depend on sample size?

Answers

a) The mean and standard deviation of the sampling of x can be determined as follows : Mean μx = Σx/n where Σx is the total of all 40 observations and n = 40 is the sample size.

The probability distribution of the population is not required for this calculation. The sum of the probabilities of all possible events is always 1. Therefore, the sum of the population proportions should be 1:30% + 20% + 20% + 30% = 100% = 1In order to determine the value of Σx for the population, the following formula may be used:Σx = (0.3)(1) + (0.2)(2) + (0.2)(3) + (0.3)(4) = 2.7So, the mean of the sampling distribution is:μx = Σx/n = 2.7/40 = 0.0675Similarly, the standard deviation of the sampling distribution is:σx = sqrt [ Σ ( xi - μx )2 / n ] = sqrt [ Σ (pi) (xi - μx)2 ] = sqrt (0.0129) = 0.1135Therefore, the mean of the sampling distribution is 0.0675 and the standard deviation is 0.1135. b) The shape of the sampling distribution of x is normal. This result is a consequence of the central limit theorem. According to the central limit theorem, when the sample size is sufficiently large, regardless of the shape of the population, the distribution of the sample means will follow an approximately normal distribution.

Hence, in this case, since the sample size is 40 which is greater than 30, the sampling distribution of x is normally distributed. The answer does not depend on sample size.

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An article in the Son jose Mercury News stated that students in the California state university system take 6 years, on average, to finish their undergraduate degrees. A freshman student believes that the mean time is less and conducts a survey of 38 students. The student obtains a sample mean of 5.6 with a sample standard deviation of 0.9. Is there sufficient evidence to support the student's claim at an α=0.1 significance level? Preliminary: a. Is it safe to assume that n≤5% of all college students in the local area? No Yes b. 15n≥30? Yes. No Test the claim: a. Determine the null and alternative hypotheses, Enter correct symbol and value. H 0
​ :μ=
H a
​ :μ
​ b. Determine the test statistic. Round to four decimal places. t= c. Find the p-value. Round to 4 decimals. p-value = d. Make a decision. Fail to reject the null hypothesis. Reject the null hypothesis. e. Write the conclusion. There is sufficient evidence to support the claim that the mean time to complete an undergraduate degree in the California state university system is less than 6 years. There is not sufficient evidence to support the claim that that the mean time to complete an undergraduate degree in the California state university system is less than 6 years.

Answers

No Test the claim Determine the null and alternative hypotheses, Enter correct symbol and value. H0:μ=6H1:μ<6b. Determine the test statistic. Round to four decimal places.

t=(x¯−μ)/(s/√n)

=(5.6-6)/(0.9/√38)

= -2.84c.

Find the p-value. Round to 4 decimals. We will use the t-distribution with degrees of freedom There is sufficient evidence to support the claim that the mean time to complete an undergraduate degree in the California state university system is less than 6 years.

df = n-1

= 38 - 1

= 37.

The area to the left of -2.84 is 0.0049. Hence, the P-value is P(t < -2.84) = 0.0049d. Make a decision. Fail to reject the null hypothesis. Reject the null hypothesise. Write the conclusion. There is sufficient evidence to support the claim that the mean time to complete an undergraduate degree in the California state university system is less than 6 years.

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Other Questions
.Presented below is information related to Oriole Company.1. On July 6, Oriole Company acquired the plant assets of Doonesbury Company, which had discontinued operations. The appraised value of the property is:Land$500,000Buildings1,500,000Equipment1,000,000Total$3,000,000Oriole Company gave 12,400 shares of its $100 par value common stock in exchange. The stock had a market price of $168 per share on the date of the purchase of the property.2. Oriole Company expended the following amounts in cash between July 6 and December 15, the date when it first occupied the building. (Prepare consolidated entry for all transactions below.)Repairs to building$304,500Construction of bases for equipment to be installed later391,500Driveways and parking lots353,800Remodeling of office space in building, including new partitions and walls466,900Special assessment by city on land52,2003. On December 20, the company paid cash for equipment, $754,000, subject to a 2% cash discount, and freight on equipment of $30,450.Prepare entries on the books of Oriole Company for these transactions. (Round intermediate calculations to 5 decimal places, e.g. 1.25124 and final answer to 0 decimal places e.g. 58,971. Credit account titles are automatically indented when amount is entered. Do not indent manually. If no entry is required, select No Entry for the account titles and enter 0 for the amounts.) An unknown compound contains only C , H , and O . Combustion of 9.50 g of this compound produced 23.2 g CO2 and 9.49 g H2O . What is the empirical formula of the unknown compound? Insert subscripts as needed. empirical formula: A put option is available for British pound with an exercise price of EUR0.7192 and a premium of EUR0.0111. Assume that there are no brokerage fees. The future spot rate at which of the option breaks even is EUR a. the seller; EUR0.7081 b. both the buyer and the seller, EUR0.7303 c. the buyer; EUR0.7303 d. the seller; EUR0.7303 e. both the buyer and the seller; EUR0.7081 f. The correct answer is not present in the other listed choices g. the buyer; EUR0.7081 On April 12, 2020, Prism Ltd., a camera lens manufacturer, paid cash of $569,000 for real estate plus $30,900 cash in closing costs. The real estate included land appraised at $182,700; land improvements appraised at $73,080; and a building appraised at $353,220.(a)(b)(c)PPE AssetAppraisedValuesRatio of Individual Appraised Value toTotal Appraised Value(a) Total Appraised ValueCost Allocation(b) Total Actual CostPresent the journal entry to record the purchase. (Do not round intermediate calculations. Round the final answers to the nearest whole dollar.) Given your chosen Macro-economic variables (please use one measuring consumer spending and one measuring business spending), briefly explain how each can be used to gauge general U.S. economic growth/activity. Also, how can each indicator be a proxy for the anticipate direction of interest rates tied to your firms cost of capital? Why is monitoring this measure of economic activity important to your personal and business lending choices...especially with regard to your firms growth and profitability (NPV...and EPS growth)...and its ultimate valuation? Why was Isabel reading the names from the first Describe the type of fraud and the impact of the fraud triangleto KBR and Halliburton FCPA (2009). Assume the credit terms offered to your firm by your suppliers are 4/10, net 60 . Calculate the cost of the trade credit if your firm does not take the discount and pays on day 60 . (Hint: Use a 365-day year.) The cost of trade credit is \%. (Rounded to two decimal places.) 1.It is found that 8% of the bottles produced by a facto...1.It is found that 8% of the bottles produced by a factory are defective . Find the probability that exactly 4 bottles are defective in a sample of 20 bottles.a)0.0522b)0.0532c)0.0525 d)0.0512 McDonalds Horizontal Analysis: Cash & Equivalents increased in dollars and percentage from 2020 to 2021: Round final answer to dollars WITH one decimal, including zero, and round final answer to whole percentage. Ex: $1,234.5; 65%; or Ex: Ex: $1,234.0; 65%A. $1,260.1 and 37%B. and C. and D. $3,449.1 and 37%McDonalds Trend Analysis with base year 2019: Total Operating Costs & Expenses increased in dollars from 2019 to 2021: Round final answer to dollars WITH one decimal, including zero. Ex: $1,234.5 or Ex: Ex: $1,234.0A. $572.3B. C. D. $410.8 The following information is available for Orset, a sole trader who does not keep full accounting records: $ Inventory 1 July 20X4138,600 30 June 20X5149,100 Purchases made for year ended 30 June 20X5 716,100Orset makes a standard gross profit of 30% on sales. Based on these figures, what is Orset's sales figure for the year ended 30 June 20X5? a. $2,352,000 b. $1,038,000 c. $917,280 d. $1.008.000 On January 1, your company repurchased 100 shares of $1 par value common stock for $5,000. On January 31, the board of directors authorized repurchase of 50 shares of the treasury stock for $60 per share. Which of the following journal entries should be recorded for January 31?O A credit to Common Stock for $2,500O A credit to Treasury Stock for $2,500O A credit to Cash for $3,000O A debit to Additional Paid-in Capital from Treasury Stock for $500O None of the entries is recorded correctly. STRATEGY IMPLEMENTATION. (How are you going to do what you want to do? Where will the company obtain the $$, the resources, the people, etc. Discuss each major departments specific duties in implementation of this strategy (management, marketing, R&D/engineering, accounting, HRM, production, MIS, finance, legal). Find solutions for your homework Find solutions for your homework businessaccountingaccounting questions and answersuse the following informatlon to prepare the september cash budget for pto company. ignore the "loan activity" section of the budget a. beginning cash balance, september 1,$50,000. b. budgeted cash recelpts from september sales, $263,000. c. direct materials are purchased on credit. purchase amounts are august (actual), $73,000; and september (budgeted), Question: Use The Following Informatlon To Prepare The September Cash Budget For PTO Company. Ignore The "Loan Activity" Section Of The Budget A. Beginning Cash Balance, September 1,$50,000. B. Budgeted Cash Recelpts From September Sales, $263,000. C. Direct Materials Are Purchased On Credit. Purchase Amounts Are August (Actual), $73,000; And September (Budgeted), student submitted image, transcription available below Show transcribed image text Expert Answer 100% 1st step All steps Final answer Step 1/1 PTO company View the full answer answer image blur Final answer Transcribed image text: Use the following informatlon to prepare the September cash budget for PTO Company. Ignore the "Loan activity" section of the budget a. Beginning cash balance, September 1,$50,000. b. Budgeted cash recelpts from September sales, $263,000. c. Direct materials are purchased on credit. Purchase amounts are August (actual), $73,000; and September (budgeted), $109,000. Payments for direct materials follow: 65% in the month of purchase and 35% In the first month after purchase. d. Budgeted cash payments for direct labor in September, $38,000 e. Budgeted depreclation expense for September, $3,300. f. Budgeted cash payment for dividends in September, $57,000. 9. Budgeted cash payment for income takes in September, \$10,700. h. Budgeted cash payment for loan interest in September, $1,800. If ABC Corporation's sales and average operating assets remain the same, ABC's return on investment will: increase if net operating income increases. increase if margin decreases. decrease if net operating income decreases. O decrease if margin increases. Security markets exist to aid the allocation of capital among which of the following?A. HouseholdsB. CorporationsC. Governmental unitsD. All of the listed answers are correct.Q. Which of the following best describes the role of an electronic communication network (ECN)?A. A network that utilizes satellite technology to ensure efficient markets.B. Trading systems that automatically match buy and sell orders at specific prices via computers.C. Networks that link international exchanges to the government's regulatory agencies.D. A system of organized rules that are designed to ensure validity of security trades.Q. If stock markets are efficient, it's difficult for investors to select portfolios of common stocks that can outperform the stock market in general.A. TrueB. FalseQ. Which legislative act specifies additional regulation of hedge funds, derivatives, credit cards, mortgages and other financial products.A. The Securities and Derivatives Act of 2008B. The Securities Act of 1933C. The Securities Act of 1934D. The Dodd-Frank Wall Street Reform and Consumer Protection Act The "Freshman 15" refers to the belief that college students gain 15 lb (or 6.8 kg) during their freshman year. Listed in the accompanying tableare weights (kg) of randomly selected male college freshmen. The weights were measured in September and later in April. Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Complete parts (a) through (c). ... Question content area top right Part 1 September 66 65 94 93 56 71 61 67 69 April71 71 105 88 53 69 60 67 69 Question content area bottom Part 1 a. Use a0.05 significance level to test the claim that for the population of freshman male college students, the weights in September are less than the weights in the following April. In this example, d is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the April weight minus the September weight. What are the null and alternative hypotheses for the hypothesis test? H0: d equals= 00 kg H1: d greater than> 00 kg (Type integers or decimals. Do not round.) Part 2 Identify the test statistic. t=enter your response here (Round to two decimal places as needed.) If the units of " are Length Time: what must be the units of w in the DE above? (a) Find the general solution to the differential equation. d'r dt2 +wz=0. (b) If the units of " Length Time what must be the units of w in the DE above? (e) If sin(t) has a period of 2r, then what must be the period of sin(at)? Eind the solution of the given initial value problem: \[ y^{*}+y^{\prime}-\sec (2), y(0)-9, y^{\prime}(0)-3, y^{\prime}(0)-2 \] Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x,y)(0,0)x 2+y 2+648x 2+y 2