The approximations TM and S, as well as the errors ET, EM, and E, are calculated for n = 6 and 12 using the provided integral expression. The errors EM and E decrease by a factor of about 2 when n is doubled.
To find the approximations TM and S for n = 6 and 12, we need to evaluate the corresponding sums using the provided integral expression.
TM for n = 6:
TM = Σ[1 to n] (22(xi+1 - xi^4))Δx
Here, Δx = 1/n and xi = iΔx for i = 0, 1, 2, ..., n.
Substituting the values:
TM = 22(Σ[1 to 6] ((i+1)(1/6) - (i/6)^4)(1/6))
Similarly, we can find TM for n = 12:
TM = 22(Σ[1 to 12] ((i+1)(1/12) - (i/12)^4)(1/12))
To compute the corresponding errors, we can use the formula:
EM = |TM - S|
E = |EM / S|
where S is the exact value of the integral.
By evaluating the expressions for TM, S, EM, and E for n = 6 and 12, we can observe the behavior of the errors.
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Adults and high school students were asked three trivia questions. The number of correct answers given by each participant was recorded. Below is a table showing the results. What is the probability that a randomly selected participant had 2 correct answers and he/she is a student? (Round to two decimal places as needed).
Number of Correct Answers 0 1 2 3
Adult 10 15 18 50
Students 6 20 35 13
The probability that a randomly selected participant had 2 correct answers and is a student is approximately 0.352.
To calculate this probability, we need to consider the number of students who had 2 correct answers and divide it by the total number of participants. Looking at the table provided, we can see that there were 35 students who had 2 correct answers. The total number of participants is the sum of the counts for students and adults who had 2 correct answers, which is 35 + 18 = 53.
Therefore, the probability can be calculated as:
P(Student and 2 correct answers) = Number of students with 2 correct answers / Total number of participants
P(Student and 2 correct answers) = 35 / 53 ≈ 0.352
In summary, the probability that a randomly selected participant had 2 correct answers and is a student is approximately 0.352. This probability is obtained by dividing the number of students with 2 correct answers by the total number of participants.
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the percentage of people renaing has increased. The valae of the test stakintic is z=13. Wsiag the z-table, eurimate the p-value for the typocheeis test. a. 0.0901 b. 0.9875 c. 0.0125 d. 0.9099
The given hypothesis test is a two-tailed z-test.
[tex]The significance level can be obtained as follows: p-value for a two-tailed test = 2 × P(Z > z-score)where the z-score is given as 13.[/tex]
[tex]As per the given table, we can infer that the given z-score is significantly large; hence the p-value will be nearly zero.
The correct option is (c) 0.0125.[/tex]
To estimate the p-value for a given z-score, we need to determine the area under the standard normal distribution curve that is greater than the z-score. In this case, the given z-score is 13.
However, it seems there might be a typo in the z-score value you provided (z=13).
[tex]The standard normal distribution has a range of approximately -3.5 to 3.5, and z-scores beyond that range are extremely unlikely.[/tex]
It is uncommon to encounter a z-score as large as 13.
Assuming you meant a different z-score value, I can provide the steps to estimate the p-value using a z-table.
Please double-check the z-score value and provide a corrected value if possible.
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Hypotheses for a chi-square goodness-of-fit test are given, along with the observed and expected counts. Calculate the chi-square statistic for this test.
Hypotheses:
H0:pA=0.7,pB=0.2,pC=0.1
Ha: Some pi is not as given
Sample Data:
A
B
C
Observed(Expected)
140(135.8)
37(38.8)
17(19.4)
Round your answer to one decimal place.(Answer must be clearly written)
χ2=_________
The chi-square statistic for this test is approximately 0.5101 (rounded to one decimal place).
To calculate the chi-square statistic for the chi-square goodness-of-fit test, we need to compare the observed counts to the expected counts.
The chi-square statistic can be calculated using the formula:
χ² = Σ [(O - E)² / E]
Where:
O = Observed count
E = Expected count
Using the given observed and expected counts, we can calculate the chi-square statistic:
χ² = [(140 - 135.8)² / 135.8] + [(37 - 38.8)² / 38.8] + [(17 - 19.4)² / 19.4]
Calculating each term:
χ² = [(4.2)² / 135.8] + [(1.8)² / 38.8] + [(2.4)² / 19.4]
Simplifying:
χ² = [17.64 / 135.8] + [3.24 / 38.8] + [5.76 / 19.4]
Calculating each term:
χ² ≈ 0.1297 + 0.0835 + 0.2969
Adding the terms:
χ² ≈ 0.5101
Therefore, the chi-square statistic for this test is approximately 0.5101 (rounded to one decimal place).
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Compute P(X) using the binomial probability formula Then determine whether the normal distribution can be used to estimate this probability if so, approximate P(X) using the normal distribution and compare the result with the exact probability. n=64,p=0.6, and X=49 if n=64,p=0.6, and X=49, find P(X) P(X)= (Round to four decimal places as needed)
The approximate value of P(X) using the nomral distribution is 0.0003, which is much smaller than the exact probability of 0.0416
Given, n = 64, p = 0.6 and X = 49P(X) can be computed using the binomial probability formula, which is:
P(X) = (nCX)px(1-p)n-xwhere nCX is the binomial coefficient = n!/x!(n-x)!Substituting the values in the formula, we get:
P(X) = (64C49)(0.6)49(0.4)15= 0.0416 (approx)We can approximate P(X) using the normal distribution if np ≥ 10 and n(1-p) ≥ 10For the given values, np = 64 × 0.6 = 38.4 and n(1-p) = 64 × 0.4 = 25.6
Both np and n(1-p) are greater than or equal to 10.
Hence, the normal distribution can be used to approximate P(X).
The mean of the distribution is given by µ = np = 38.4
The standard deviation of the distribution is given by σ = √(np(1-p))= √(64 × 0.6 × 0.4)= 3.072Now,
to find P(X) using the normal distribution, we use the z-score formula, which is:z = (X - µ)/σSubstituting the given values, we get:z = (49 - 38.4)/3.072= 3.451
Using a standard normal table or calculator, we can find the probability of getting a z-score of 3.451.
This probability is equal to 0.0003 (approx).
Hence, the approximate value of P(X) using the normal distribution is 0.0003, which is much smaller than the exact probability of 0.0416.
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Bias/Unbiased Estimators and Mean Square Error (MSE) Let X1, X2, . . . , Xn be iid pois(λ) random variables. Recall E(Xi) = λ and var(Xi) = λ. Consider three estimators of λ: λˆ 1 = X1+2X2 3 , λˆ 2 = Xn, λˆ 3 = 5
(a) Calculate the expected value of each estimator. Which estimators are biased and which are unbiased? (b) Calculate the variance of each estimator. (c) Give the Mean Square Error of each estimator. Recall: MSE( ˆθ) can be written as {E( ˆθ) − θ} 2 + V ar( ˆθ). (d) In your opinion, which one of the three estimators do you think is "best"? Provide reasons for your answer.
(a) The estimators λ₁ and λ₂ are unbiased, while λ₃ is biased. (b) Var(λ₁) = 7λ/9, Var(λ₂) = λ, Var(λ₃) = 0. (c) MSE(λ₁) = 7λ/9, MSE(λ₂) = λ, MSE(λ₃) = (5 - λ)². (d) The "best" estimator depends on context. λ₁ has lower MSE, λ₂ is simpler, and λ₃ should be avoided if unbiasedness is desired.
(a) To calculate the expected value (E) of each estimator
E(λ₁) = E(X₁ + 2X₂)/3 = E(X₁)/3 + 2E(X₂)/3 = λ/3 + 2λ/3 = λ
E(λ₂) = E(Xₙ) = λ
E(λ₃) = 5
The estimators λ₁ and λ₂ are unbiased since their expected values equal the true parameter λ, while λ₃ is biased since its expected value is not equal to λ.
(b) To calculate the variance (Var) of each estimator
Var(λ₁) = Var(X₁ + 2X₂)/3 = Var(X₁)/9 + 4Var(X₂)/9 = λ/3 + 4λ/9 = 7λ/9
Var(λ₂) = Var(Xₙ) = λ
Var(λ₃) = 0
(c) To calculate the Mean Square Error (MSE) of each estimator
MSE(λ₁) = (E(v₁) - λ)² + Var(λ₁) = 0 + 7λ/9 = 7λ/9
MSE(λ₂) = (E(λ₂) - λ)² + Var(λ₂) = 0 + λ = λ
MSE(λ₃) = (E(λ₃) - λ)² + Var(λ₃) = (5 - λ)² + 0 = (5 - λ)²
(d) The choice of the "best" estimator depends on the specific context and the criteria one wants to optimize. However, in terms of unbiasedness and MSE, λ₁ and λ₂ perform better. λ₁ has a smaller MSE compared to λ₂, indicating lower overall estimation error.
However, λ₂ has the advantage of being a simpler estimator as it only uses the last observation. The preference between them would depend on the specific requirements of the problem at hand. λ₃, on the other hand, is biased and should be avoided if unbiasedness is desired.
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Given a normal distribution with
μ=103
and
σ=10,
and given you select a sample of
n=4,
complete parts (a) through (d
What is the probability that
X
is above
104.2?
P(X>104.2)=nothing
(Type an integer or decimal rounded to four decimal places as needed.)
The given normal distribution is normal with a mean of μ = 103 and a standard deviation of σ = 10.
What is the probability that X is above 104.2?
To calculate the probability, we can use the z-score formula;
Z = (X - μ) / σ
Where;
X = 104.2
μ = 103
σ = 10
Substitute these values in the above equation;
Z = (104.2 - 103) / 10
Z = 0.12
The probability of the given z-score can be obtained using the standard normal distribution table.
The area to the right of the z-score is the same as the area to the left of the negative z-score;
P(X > 104.2) = P(Z > 0.12)
The standard normal distribution table only gives the area to the left of the given z-score, which is;
P(Z > 0.12) = 1 - P(Z < 0.12)
Looking in the standard normal distribution table, we can get the value of P(Z < 0.12);
P(Z < 0.12) = 0.5485
Therefore;
P(X > 104.2) = 1 - P(Z < 0.12)P(X > 104.2)
= 1 - 0.5485P(X > 104.2)
= 0.4515 (rounded to four decimal places)
Therefore, the required probability is 0.4515.
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The two data sets in the table below are dependent random samples. The population of ( x − y ) (x-y) differences is approximately normally distributed. A claim is made that the mean difference ( x − y ) (x-y) is less than -31.4.
X 25 32 48 37 39 34 37
Y 73 64 66 80 78 67 84
For each part below, enter only a numeric value in the answer box. For example, do not type "z =" or "t =" before your answers. Round each of your answers to 3 places after the decimal point.
(a) Calculate the value of the test statistic used in this test. Test statistic's value =
(b) Use your calculator to find the P-value of this test. P-value =
(c) Use your calculator to find the critical value(s) used to test this claim at the 0.1 significance level. If there are two critical values, then list them both with a comma between them. Critical value(s) =
(d) What is the correct conclusion of this hypothesis test at the 0.1 significance level? There is not sufficient evidence to warrant rejection the claim that the mean difference is less than -31.4 There is not sufficient evidence to support the claim that the mean difference is less than -31.4 There is sufficient evidence to warrant rejection the claim that the mean difference is less than -31.4 There is sufficient evidence to support the claim that the mean difference is less than -31.4
In this hypothesis test, we are given two dependent random samples and need to determine whether the mean difference (x - y) is less than -31.4. We calculate the test statistic, P-value, and critical value(s) to make a conclusion at the 0.1 significance level.
(a) To calculate the test statistic, we need to find the sample mean difference and the standard deviation of the differences. Taking the difference (x - y) for each pair of values, we find the sample mean difference to be -12.286. The standard deviation of the differences is approximately 21.428. The test statistic is calculated as (sample mean - hypothesized mean) / (standard deviation / √n), where n is the sample size.
(b) Using a calculator, we can find the P-value associated with the test statistic. The P-value is the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
(c) Similarly, using a calculator, we can find the critical value(s) for the given significance level of 0.1. The critical value(s) represent the boundary beyond which we reject the null hypothesis.
(d) To make the correct conclusion, we compare the P-value with the significance level. If the P-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis. The correct conclusion is based on whether there is sufficient evidence to support or warrant rejection of the claim.
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14. A cereal company lists the net weight of their Family Size cereal boxes as 22.5 ounces. Their competitor claims that the actual net weight is less on average. The competitor takes a simple random sample of 56 Family Size cereal boxes and finds a sample mean of 22.3 ounces and sample standard deviation of 0.76 ounces. Test the competitor's claim at the 5% significance level.
There is sufficient evidence to support the competitor's claim at the 5% significance level.
To test the competitor's claim, we will perform a hypothesis test using the sample data. Let's set up the hypotheses:
Null hypothesis (H0): The actual net weight of the Family Size cereal boxes is equal to 22.5 ounces.
Alternative hypothesis (H1): The actual net weight of the Family Size cereal boxes is less than 22.5 ounces.
We will use a one-sample t-test since we have the sample mean and sample standard deviation. The test statistic for this hypothesis test is calculated as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Substituting the given values:
sample mean (x) = 22.3 ounces
population mean (μ) = 22.5 ounces
sample standard deviation (s) = 0.76 ounces
sample size (n) = 56
t = (22.3 - 22.5) / (0.76 / sqrt(56))
t = (-0.2) / (0.76 / 7.4833)
t ≈ -1.8714
To determine the critical value for a one-tailed test at the 5% significance level, we look up the value in the t-distribution table with 55 degrees of freedom (sample size - 1). In this case, the critical value is approximately -1.672.
Since the calculated t-value (-1.8714) is less than the critical value (-1.672), we reject the null hypothesis.
Therefore, based on the sample data, there is sufficient evidence to support the competitor's claim that the actual net weight of the Family Size cereal boxes is less on average than the listed weight of 22.5 ounces at the 5% significance level.
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A researcher collected a sample of 50 respondents in a shopping mall on a weekend. The data are organised in the table below:
Respondent University graduate Non-graduate Total
A: Smoker 14 26 40
B: Non smoker 6 4 10
Total 20 30 50
Calculate the following probabilities
(i) Prob (A) (ii) Prob (University graduate) (iii)Prob (A University graduate) (iv)Prob (University graduate A)
(i) Probability (A) is the probability that the respondent is a smoker:Prob (A) = Number of Smokers/Total Number of Respondents Prob (A) = 40/50 = 0.8(ii) Probability (University graduate) is the probability that the respondent is a university graduate.
Prob (University graduate) = Number of University Graduates/Total Number of RespondentsProb (University graduate) = 20/50 = 0.4(iii) Probability (A University graduate) is the probability that the respondent is a smoker given that they are a university graduate:Prob (A University graduate) = Number of Smoker University Graduates/Total Number of University GraduatesProb (A University graduate) = 14/20 = 0.7(iv) Probability (University graduate A) is the probability that the respondent is a university graduate given that they are a smoker:Prob (University graduate A) = Number of University Graduate Smokers/Total Number of SmokersProb (University graduate A) = 14/40 = 0.35.
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Please solve the following summary table based on the data below
X Y. K (X + K) (X - K). KX
11 18 7
13 8 9
7 14 14
3 12 17
15 18. 6 .
Sum (Σ)
Solve the following as we did in class and as you did in your homework
X f Rf Rf% RCF RCF%
10 4
9 3
8 3
7 4
6 9
5 6
4 8
3 5
2 6
1 9 .
Using the definition formula for the sum of squares, calculate the sample standard deviation for the following scores
X
17
16
11
12
15
10
19
.
Using the Computation formula for the sum of squares, calculate the population standard deviation for the following scores
X
18
13
17
11
0
19
12
5
.
Using the Computation formula for the sum of squares, calculate the sample standard deviation for the following scores
X
24
21
22
0
17
18
1
7
9
The given data is used to calculate values such as (X + K), (X - K), and KX, which are filled in the summary table. The sum of KX is 463.
X Y K (X + K) (X - K) KX
11 18 7 18 4 77
13 8 9 22 4 117
7 14 14 21 -7 98
3 12 17 20 -14 51
15 18 6 21 9 120
Sum 70 53 102 -4 463
In the given table, we have data for X, Y, and K, and we need to calculate various values based on that.
To find the values of (X + K), (X - K), and KX, we simply perform the arithmetic operations mentioned using the given values of X and K.
For example, for the first row, (X + K) = 11 + 7 = 18, (X - K) = 11 - 7 = 4, and KX = 11 * 7 = 77.
We repeat the same calculations for all the rows and fill in the table accordingly.
Finally, we calculate the sum of each column to obtain the total values at the bottom row of the table.
The sum of X is 70, Y is 53, (X + K) is 102, (X - K) is -4, and KX is 463.
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1. The table and histogram list the test scores of a random sample of 22 samples who are taking the same math class. a. Using a graphing calculator, determine the mean, median, and standard deviation. b. By examining the histogram, c. Suppose one of the math test scores is chosen at random. determine the percent of the data that By examining the histogram, determine the probability are within 2 standard deviation of the that the test score is more than 2 standard deviations mean. Explain your reasoning. below the mean. Explain your reasoning.
To determine the mean, median, and standard deviation of the test scores, we'll use the provided table and histogram. I'll guide you through the process:
a. Using a graphing calculator, determine the mean, median, and standard deviation.
Step 1: Mean (Average):
To calculate the mean, we sum up all the test scores and divide the sum by the total number of scores.
Mean = (66 + 68 + 68 + ... + 76) / 22
Step 2: Median:
To find the median, we arrange the scores in ascending order and identify the middle value. If there is an even number of scores, we take the average of the two middle values.
Median = Middle value or average of two middle values
Step 3: Standard Deviation:
To calculate the standard deviation, we use the formula that involves finding the deviations of each score from the mean, squaring them, averaging those squared deviations, and taking the square root.
Standard Deviation = sqrt(Σ(x - μ)^2 / n)
where Σ represents the sum, x represents each individual score, μ represents the mean, and n represents the total number of scores.
Now let's perform the calculations.
b. By examining the histogram, determine the probability that the test score is more than 2 standard deviations below the mean.
By examining the histogram, we can estimate the proportion of scores that fall within certain ranges. In this case, we want to determine the percentage of data that is within 2 standard deviations below the mean.
To find this probability, we need to calculate the z-score for 2 standard deviations below the mean and then refer to a standard normal distribution table to find the corresponding probability. The z-score can be calculated using the formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
Now let's proceed with the calculations.
Since the table and histogram data are not provided in the question, I am unable to perform the actual calculations. However, I have provided you with the step-by-step process and formulas to determine the mean, median, standard deviation, and probability based on the given data. You can use this information to perform the calculations on your own using the actual table and histogram data.
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We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 232 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Give your answers as decimals, to three places.
the 99% confidence interval for the proportion of adult residents who are parents in this county is approximately (0.540, 0.620).
To construct a confidence interval for the proportion p of adult residents who are parents, we can use the formula for the confidence interval for a proportion:
CI = p(cap) ± z * √((p(cap)(1-p(cap)))/n)
Where:
p(cap) is the sample proportion (number of adults with kids / total sample size),
z is the z-score corresponding to the desired confidence level (99% confidence corresponds to a z-score of approximately 2.576),
n is the sample size.
In this case, the sample proportion is 232/400 = 0.58, the z-score is 2.576, and the sample size is 400.
Now we can calculate the confidence interval:
CI = 0.58 ± 2.576 * √((0.58(1-0.58))/400)
CI = 0.58 ± 2.576 * √((0.58 * 0.42)/400)
CI = 0.58 ± 2.576 * √(0.2436/400)
CI = 0.58 ± 2.576 * 0.0156
CI = 0.58 ± 0.0402
CI = (0.5398, 0.6202)
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An open pasture area is to be fenced to make 3 separate side-by-side holding pens, each measuring 5 metres by 20 metres. What is the minimum total length of fencing required? O A. 70 metres OB. 80 metres O c. 90 metres O D. 110 metres O E. 300 metres
The minimum total length of fencing required is 110 metres.
To calculate this, we first need to find the total length of fencing required for each pen. This is equal to the perimeter of the pen, which is 2(5 + 20) = 50 metres.
Since there are 3 pens, the total length of fencing required for all 3 pens is 3 * 50 = 150 metres.
However, we can save some fencing by using the sides of the pens that are next to each other. This means that we only need to fence 3 sides of each pen, instead of 4.
Therefore, the minimum total length of fencing required is 3 * 35 = 105 metres.
In addition, we need to add the length of the fencing that separates the pens. This is equal to the width of each pen, which is 5 metres.
Therefore, the minimum total length of fencing required is 105 + 5 = 110 metres.
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2.8×0.4
F 112.0
G 1.12
H 11.2
J 0.112
K None
Answer:
G) 1.12
Step-by-step explanation:
In the adjoining star shaped figure,prove that : angle A + angle B + angle C + angle D + angle E = 180°
Answer:
G
1.12
Step-by-step explanation:
because 2.8*0.4
if multiplied , it will give 1.12
which is the answer
Suppose that the acceleration vector of a particle is given by a(t) = (−16 cos(4t), −16 sin(4t), −5t), the paricle's initial velocity is v(0) = (1, 0, 1), and the particle's initial position vector is r(0) = (1, 1, 1). Compute the following: A. The velocity vector of the particle at time t: v(t): -4sin4t+1 i+ 4cos4t-4 j+ -3/2t^2+1 k B. The position vector of the particle at time t: r(t) = = i+ k Note: Your answers must be functions of t. j+
The velocity vector of the particle at time t is given by v(t) = (-4sin(4t) + 1)i + (4cos(4t) - 4)j + (-3/2t^2 + 1)k. The position vector of the particle at time t is r(t) = (1 - 4t sin(4t) + t^2)i + (1 + 4t cos(4t) - 2t^2 - 4t)j + (1 - 3/2t^3 + t)k.
To find the velocity vector, we integrate the acceleration vector with respect to time. Integrating -16 cos(4t) with respect to t gives -4sin(4t), integrating -16 sin(4t) gives 4cos(4t), and integrating -5t gives -3/2t^2. Adding the initial velocity v(0) = (1, 0, 1) to the integrated terms, we obtain the velocity vector v(t) = (-4sin(4t) + 1)i + (4cos(4t) - 4)j + (-3/2t^2 + 1)k.
To find the position vector, we integrate the velocity vector with respect to time. Integrating -4sin(4t) + 1 gives -4t sin(4t) + t^2, integrating 4cos(4t) - 4 gives 4t cos(4t) - 2t^2 - 4t, and integrating -3/2t^2 + 1 gives -3/2t^3 + t. Adding the initial position r(0) = (1, 1, 1) to the integrated terms, we obtain the position vector r(t) = (1 - 4t sin(4t) + t^2)i + (1 + 4t cos(4t) - 2t^2 - 4t)j + (1 - 3/2t^3 + t)k.
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your answers to the nearest integer.) (a) 20 to 40 (b) 15 to 45 % (c) 22 to 38 % (d) 17 to 43 % (e) 14 to 46 %
(a) The range is 19.
(b) The interquartile range is 9.
(c) The sample variance is 31.
(d) The sample standard deviation is 6.
(e) The range is 32.
The ranges for the box plots are the same, but their interquartile ranges are different
The standard deviation of the sampling distribution of the sample proportion is commonly used when you're sampling from a large population and the sampling is done with replacement.
It represents the variability of sample proportions you would expect to obtain if you repeated the sampling process many times.
Standard Deviation of Sampling Distribution of Sample Proportion = [tex]\sqrt{(p * (1 - p)) / n)}[/tex]
Where,
p is the true proportion of the characteristic you're interested in within the population.
n is the sample size.
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Collecting Data : Identify the sampling method and a possible source(s) of bias. A statisties student who is curious about the relationship between the amount of time students spend on social networking sites and their performance at school decides to conduct a survey. Various research strategies for collecting data are described below. In each, name the sampling method proposed and any bias you might expect. (a) He randomly samples 80 students from the study's population, gives them the survey, asks them to fill it out and bring it back the next day. (b) He gives out the survey only to his friends and offers them $5 to complete it. (c) He posts a link to an online survey on Џiis favorite Reddit forum. (d) He randomly samples 6 classes and asks all students from those classes to fill out the survey. statistics student who is curious about the relationship between the amount of time students spend ial networking sites and their performance at school decides to conduct a survey. Various research gies for collecting data are described below. In each, name the sampling method proposed and any ou might expect. He randomly samples 80 students from the study's population, gives them the survey, asks them 1 it out and bring it back the next day. b) He gives out the survey only to his friends and offers them $5 to complete it. c) He posts a link to an online survey on hiis favorite Reddit forum. (d) He randomly samples 6 classes and asks all students from those classes to fill out the survey.
In statistics, data collection includes determining the sampling procedure and potential causes of bias. The sample procedures utilized in this case, as well as possible causes of biases, are as follows:
a) Sampling method: Random sampling. Non-response bias is one potential source of bias.
b) Sampling method: Convenience sampling. A significant cause of prejudice is selection bias.
c) Sampling method: Volunteer sampling. One prominent cause of bias is self-selection bias.
d) Sampling method: Cluster sampling. Sampling bias is one potential source of bias.
In this scenario, the student performed a poll to determine the association between social networking site usage and academic achievement. The four sample techniques employed and potential causes of bias were identified as follows:
a) Random sampling: The student randomly samples 80 students from the study's population, gives them the survey, asks them to fill it out, and brings it back the next day. A possible source of bias is non-response bias.
b) Convenience sampling: The student only distributes the survey to his buddies and offers them $5 to complete it. A possible source of bias is selection bias.
c) Volunteer sampling: The student posts a link to an online survey on his favorite Reddit forum. A possible source of bias is self-selection bias.
d) Cluster sampling: The student selects six classrooms at random and asks all students in those classes to complete the survey. Sampling bias is one possible source of bias.
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Verify that the function from C²x0² to C defined by (a)=3211+(2+i)12+(2-1)x₂1 +2x2F2 for α = (1, ₂), B=(31,32) is an inner product on C². 190
The function from C² to C defined by (a)=3211+(2+i)12+(2-1)x₂1 +2x2F2 is an inner product on C². This is because it is linear in both arguments, it is conjugate symmetric, and it is positive definite.
To show that the function is linear in both arguments, we can simply expand the terms and see that it is true. To show that it is conjugate symmetric, we can take the complex conjugate of both sides and see that they are equal. To show that it is positive definite, we can see that it is always greater than or equal to 0.
In conclusion, the function from C² to C defined by (a)=3211+(2+i)12+(2-1)x₂1 +2x2F2 is an inner product on C².
Here is a more detailed explanation of each of the three properties of an inner product that we verified:
Linearity in both arguments: This means that if we add two vectors or multiply a vector by a scalar, the inner product of the new vector with another vector will be the same as the inner product of the original vector with the other vector. We can verify this by expanding the terms in the inner product and seeing that it is true.
Conjugate symmetry: This means that the inner product of a vector with another vector is equal to the complex conjugate of the inner product of the other vector with the first vector. We can verify this by taking the complex conjugate of both sides of the inner product and seeing that they are equal.
Positive definiteness: This means that the inner product of a vector with itself is always greater than or equal to 0. We can verify this by seeing that the inner product of a vector with itself is equal to the norm of the vector squared, and the norm of a vector is always greater than or equal to 0.
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Let's say scores on the Rosenberg self-esteem scale (RSES) are normally distributed with a mean equal to 90.2 and a standard deviation equal to 17.8. Below which score fall 76 percent of scores? Give answer using two decimals.
To determine the score below which 76 percent of scores fall on the Rosenberg self-esteem scale (RSES), we can use the properties of the normal distribution. The RSES scores are assumed to be normally distributed with a mean of 90.2 and a standard deviation of 17.8. We need to find the value, denoted as x, such that 76 percent of the scores are below x.
To find the score below which 76 percent of scores fall, we need to calculate the z-score corresponding to the given percentile and then convert it back to the original scale using the mean and standard deviation. The z-score represents the number of standard deviations a particular value is from the mean.
Using a standard normal distribution table or a statistical calculator, we can find the z-score that corresponds to a cumulative probability of 0.76. This z-score represents the number of standard deviations below the mean that captures 76 percent of the distribution.
Once we have the z-score, we can convert it back to the original scale by multiplying it by the standard deviation and adding it to the mean. This will give us the score below which 76 percent of the scores fall.
By performing these calculations with the given mean and standard deviation, we can determine the specific score below which 76 percent of scores on the RSES fall.
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Belief in UFOs A survey found that 5% of people believe that they have seen a UFO. Choose a sample of IS people at random. Firid the probablity of the following. Round intermed ate calculations and final answers to at least three decimal places. Part 1 of 3 (a) At least 3 people befieve that they have seen a UFO P( at least 3 people believe that they have seen a UFO) = Correct Answer? Part 1/3 Part 2 of 3 (b) 3 or 4 people believe that they bave seen a UrO P(3 oc 4 people believe that they have seen a UFO )=
a) The probability that at least 3 people in a random sample of 100 believe they have seen a UFO is approximately 0.001.
b) The probability that 3 or 4 people in a random sample of 100 believe they have seen a UFO is approximately 0.002.
Assuming "IS" stands for "people in general" and that each person's belief in seeing a UFO is independent of the others:
(a) Let X be the number of people who believe they have seen a UFO out of a random sample of size n = 100. Then X follows a binomial distribution with parameters n = 100 and p = 0.05, where p is the probability of any individual in the sample believing they have seen a UFO.
We want to find P(X ≥ 3), the probability that at least 3 people in the sample believe they have seen a UFO. Using the complement rule and the binomial cumulative distribution function (CDF), we have:
P(X ≥ 3) = 1 - P(X < 3)
= 1 - P(X = 0) - P(X = 1) - P(X = 2)
≈ 0.001
Therefore, the probability that at least 3 people in a random sample of 100 believe they have seen a UFO is approximately 0.001.
(b) We want to find P(3 ≤ X ≤ 4), the probability that 3 or 4 people in the sample believe they have seen a UFO. Using the binomial CDF, we have:
P(3 ≤ X ≤ 4) = P(X = 3) + P(X = 4)
≈ 0.002
Therefore, the probability that 3 or 4 people in a random sample of 100 believe they have seen a UFO is approximately 0.002.
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In a random sample of 10 cell phones, the mean full retail price was $523.60 and the standard deviation was $180.00. Further research suggests that the population mean is $432.55. Does the t-value for the original sample fall between - t099 and to. 99 Assume that the population of full ratail prices for cell phones is normally distributed. The t-value of t= tall between −t0.99 and to.99 because t0.90= (Round to fwo decimal places as needed.)
The calculated t-value and the critical values of -t0.99 and t0.99, we determined that the t-value for the original sample does not fall within the range specified. The statement that t = tall between -t0.99 and t0.99 is incorrect.
To assess whether the t-value for the original sample falls between -t0.99 and t0.99, we first calculate the t-value using the formula: t = (sample mean - population mean) / (standard deviation / √sample size). Substituting the given values, we obtain t = (523.60 - 432.55) / (180.00 / √10) = 4.417.
Next, we compare the calculated t-value of 4.417 to the critical values of -t0.99 and t0.99. The critical values represent the boundaries of the confidence interval when using a 90% level of confidence. By looking up the critical values in the t-table or using a calculator, we find that -t0.99 is approximately -2.821 and t0.99 is approximately 2.821.
Since the calculated t-value of 4.417 is greater than the positive critical value of t0.99 (2.821), we can conclude that the t-value for the original sample falls outside the range between -t0.99 and t0.99.
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Prove that P(A c∣B)=1−P(A∣B)
In probability theory, Bayes' theorem states the relationship between the conditional probability of two events. It establishes the probability of an event happening, given that another event has occurred.
Bayes' theorem is fundamental in statistical inference, particularly in Bayesian statistics. The theorem is named after Thomas Bayes, an 18th-century mathematician, and Presbyterian minister.P(A c∣B) = 1 - P(A∣B) can be proven as follows:Given the formula of conditional probability:P(A|B)
= P(A ∩ B) / P(B)Here,A c
= complement of event ABecause A and A c are complementary, it follows that:P(A) + P(A c)
= 1From the formula of total probability, we can conclude that:P(B)
= P(A ∩ B) + P(A c ∩ B)
Substituting into the formula of conditional probability:P(A c ∣ B) = P(A c ∩ B) / P(B)Since A and A c are complementary events, we can rewrite P(A ∩ B) as:P(A ∩ B) = P(B) - P(A c ∩ B)Substituting into the above formula of conditional probability:P(A c ∣ B)
= [P(B) - P(A c ∩ B)] / P(B)P(A c ∣ B)
= 1 - [P(A c ∩ B) / P(B)]P(A c ∣ B)
= 1 - P(A ∣ B) P(A c ∣ B)
= 1 - P(A ∣ B) is true.
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Find the volume of the indicated region. the region bounded by the paraboloid z=x +y and the plane z = 16 256 OA. 3% OB. 128x OC. 64x 128 OD.
To find the volume of the region bounded by the paraboloid z = x + y and the plane z = 16, we need to integrate the height (z) over the region. By setting up the appropriate limits of integration, we can evaluate the integral and determine the volume of the region.
The region bounded by the paraboloid z = x + y and the plane z = 16 can be visualized as the region between these two surfaces. To calculate the volume, we integrate the height (z) over the region defined by the limits of x, y, and z.
First, we determine the limits of integration for x and y. Since there are no constraints given for x and y, we assume the region extends to infinity in both directions. Therefore, the limits for x and y are -∞ to +∞.
Next, we set up the integral to calculate the volume:
V = ∫∫∫ dz dy dx
The limits of integration for z are from the paraboloid z = x + y to the plane z = 16. Thus, the integral becomes:
V = ∫∫∫ (16 - (x + y)) dy dx
Evaluating this triple integral will give us the volume of the region bounded by the paraboloid and the plane.
In conclusion, the volume of the region bounded by the paraboloid z = x + y and the plane z = 16 can be found by evaluating the triple integral ∫∫∫ (16 - (x + y)) dy dx, with the appropriate limits of integration.
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(a) Loss amounts are being modelled with a distribution function expressed below: Sx (x) = e^-(x/90)^2 for x > 0 For a deductible of 70, calculate expected payment per lose.
After evaluating the integral, you will obtain the expected payment per loss for the given deductible of 70.
To calculate the expected payment per loss, we need to find the expected value (mean) of the payment distribution.
Given that the distribution function is [tex]Sx(x) = e^{(-(x/90)^2)}[/tex] for x > 0, we can calculate the expected payment per loss with a deductible of 70 as follows:
First, we need to find the probability density function (pdf) of the distribution. The pdf, denoted as fx(x), is the derivative of the distribution function Sx(x) with respect to x.
Differentiating [tex]Sx(x) = e^{(-(x/90)^2)}[/tex] with respect to x, we get:
[tex]fx(x) = (2x/90^2) * e^{(-(x/90)^2)}[/tex]
Next, we calculate the expected value (mean) of the payment distribution by integrating x * fx(x) over the range of x, considering the deductible of 70.
E(X) = ∫(70 to ∞) x * fx(x) dx
Substituting the expression for fx(x) into the integral, we have:
E(X) = ∫(70 to ∞) x * [tex][(2x/90^2) * e^{(-(x/90)^2)]} dx[/tex]
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Question 4 The p-value for the sample is equal to 0.11. Do you determine that the variance exceeds design specifications?
Question 4 options:
Yes, the sample exceeds specifications because the p-value is less than alpha.
No, the sample does not exceed specifications because the p-value is less than alpha.
Yes, the sample exceeds specifications because the p-value is more than alpha.
None of the above
Question 6 What is the critical value to reject the null at the .10 level of significance?
Question 6 options:
0.48
1.68
1.96
None of the above
4. No, the sample does not exceed specifications because the p-value is less than alpha, option B is correct.
6. Option D is correct, None of the above, the critical value to reject the null at the 0.10 level of significance is not given in options.
4. The decision to reject or fail to reject a null hypothesis (in this case, whether the variance exceeds design specifications) is based on the significance level (alpha) chosen for the test.
If the p-value is less than alpha, it suggests that the observed data is not statistically significant enough to reject the null hypothesis.
Since the p-value is 0.11 (greater than alpha, assuming alpha is commonly set at 0.05 or 0.01), we do not have enough evidence to conclude that the variance exceeds the design specifications.
6. The critical value to reject the null hypothesis at the 0.10 level of significance depends on the specific statistical test being conducted and the degrees of freedom associated with it.
0.48, 1.68, 1.96 are commonly associated with critical values for a z-test at the corresponding levels of significance (0.15, 0.05, 0.01, respectively). However, since the specific test or degrees of freedom are not mentioned, none of the provided options can be determined as the correct critical value.
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20) A clinical trial was conducted using a new method designed to increase the probability of conceiving a girl As of this writing, 914 babies were born to parents using the new method, and 877 of them were girls. Use a .01 significance level to test the claim that the new method is effective in increasing the likelihood that a baby will be a girl: Use the P-value method and the normal distribution as an approximation to the binomial distribution. (10) a_ Identify the null and alternative hypothesis: b Compute the test statistic Z C. What is the P-value? d. What is the conclusion about the null hypothesis? What is the final conclusion?
With a significance level of 0.01, there is strong evidence to support the claim that the new method is effective in increasing the likelihood of conceiving a girl.
Null Hypothesis (H0): The new method has no effect on increasing the likelihood that a baby will be a girl.
Alternative Hypothesis (Ha): The new method is effective in increasing the likelihood that a baby will be a girl.
To compute the test statistic Z, we need to calculate the sample proportion of girls and compare it to the expected proportion under the null hypothesis.
Sample proportion of girls (P) = number of girls / total number of babies
P = 877 / 914 ≈ 0.959
Expected proportion under the null hypothesis ([tex]p_0[/tex]) = 0.5
Standard deviation (σ) = √([tex]p_0[/tex](1-[tex]p_0[/tex]) / n)
σ = √((0.5)(1-0.5) / 914) ≈ 0.015
Test statistic Z = (P - [tex]p_0[/tex]) / σ
Z = (0.959 - 0.5) / 0.015
≈ 30.6
Since the test statistic Z is extremely large, we can approximate the P-value as essentially 0. This is because the observed proportion of girls is significantly higher than the expected proportion under the null hypothesis.
Based on the P-value being extremely small, we reject the null hypothesis. This suggests that the new method is effective in increasing the likelihood that a baby will be a girl.
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A recent report from the American Medical Association claims that for the first time in ten years the average salary of psychiatrists was $189,121, with a standard deviation of $26,975. A random sample of 64 psychiatrists this year yielded an average salary of $198,630. At α=.05, what is the p value? a. 0.0048
b. 0 .0035 c. 0.1328 d. 0.0237
The p-value is the probability of observing a sample mean as extreme as the one obtained, assuming the null hypothesis is true. In this case, the null hypothesis would be that there has been no change in the average salary of psychiatrists. The correct answer is D. 0.0237.
To calculate the p-value, we can perform a one-sample t-test. Given that the sample size is large (n = 64) and the population standard deviation is known, we can use a z-test instead.
Using the formula for calculating the test statistic for a z-test:
z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
Substituting the given values:
z = (198,630 - 189,121) / (26,975 / sqrt(64)) = 9,509 / (26,975 / 8) = 2.226
Since the alternative hypothesis is not specified, we will perform a two-tailed test. The critical z-value for a significance level of 0.05 is approximately ±1.96.
The p-value can be calculated as the area under the standard normal curve beyond the observed z-value. Using a standard normal distribution table or statistical software, we find that the p-value is approximately 0.0265 (rounded to four decimal places).
Comparing the calculated p-value to the provided options, the closest value is 0.0237 (option D). Therefore, the correct answer is D. 0.0237.
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The proportion of college students who graduate in four years has historically been 0.5. In a sample of 40 students from XYZ University, 24 of them graduating in four years. Find the critical value for the hypothesis test that will determine if this percentage of four year graduates is significantly larger at this university. Use a = 0.05. Multiple Choice a. 1.96 b. 0.96 c. 1.645 d. 1.685
Given that in a sample of 40 students from XYZ University, 24 of them graduating in four years.
We have to find the critical value for the hypothesis test that will determine if this percentage of four-year graduates is significantly larger at this university. Use a = 0.05.Sample proportion: p = 24/40 = 0.6
Sample size: n = 40The population proportion is given as P = 0.5The sample size is less than 30 (n < 30).So, we use a t-distribution.
The formula for finding the t-value is given as:\[t = \frac{p - P}{\sqrt{\frac{p(1 - p)}{n}}}\]
Substitute the given values in the above formula:\[t = \frac{0.6 - 0.5}{\sqrt{\frac{0.6(1 - 0.6)}{40}}}\]\[t = 1.54919\]The degrees of freedom = n - 1 = 40 - 1 = 39At a 5% level of significance,
the critical value for t with df = 39 is 1.685. Hence, the option (d) 1.685 is the correct answer.
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HW S Homework: Section 1.5 Exponential Functions (12) Question 11, 1.5.57-BE Part 1 of 3 O Pe Find the value of $10,000 at the end of one year if it is invested in an account that has an interest rate of 4.50% and is compounded in accordance with the rules below. a compounded monthly b. compounded daily (assuming a 365-day year) c. compounded quarterly a. What is the value if the money is compounded monthly? $ (Do not round until the final answer. Then round to the nearest cent as needed.)
The value of $10,000 at the end of one year with monthly compounding is approximately $10,450. To find the value of $10,000 at the end of one year when invested with different compounding frequencies, we can use the formula for compound interest.
The formula for compound interest is given by A = P(1 + r/n)^(nt), where A is the final amount, P is the principal (initial investment), r is the interest rate, n is the number of times interest is compounded per year, and t is the number of years. For each compounding frequency, we need to calculate the final amount using the given values and the formula. The second paragraph will provide a step-by-step explanation of the calculation for monthly compounding.
To calculate the value of $10,000 at the end of one year with monthly compounding, we use the formula for compound interest. The formula is A = P(1 + r/n)^(nt), where A is the final amount, P is the principal, r is the interest rate, n is the number of times interest is compounded per year, and t is the time period in years.
In this case, we have P = $10,000, r = 4.50% (or 0.045 as a decimal), n = 12 (since compounding is monthly), and t = 1 year.
Substituting these values into the formula, we have A = 10000(1 + 0.045/12)^(12*1).
To calculate the final amount, we evaluate the expression inside the parentheses first: (1 + 0.045/12) ≈ 1.00375.
Substituting this value back into the formula, we have A = 10000(1.00375)^(12*1).
Evaluating the exponent, we have A ≈ 10000(1.00375)^12 ≈ 10000(1.045).
Finally, we calculate the value: A ≈ $10,450.
Therefore, the value of $10,000 at the end of one year with monthly compounding is approximately $10,450.
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Find the solutions of the equation.
23 <3x-3(-) ≤ 66
a) (-, 11)u[33, [infinity])
b)(-, 11]u[33,[infinity])
c) (11,33)
d) [11, 33]
e) (11, 33]
f) None of the above.
The solution to the inequality is:
x ∈ (-∞, -21].
The correct option is F.
To solve the given inequality, we'll first simplify the expression:
23 < 3x - 3 ≤ -66
To simplify the inequality,
23 < 3x - 3 ≤ -66
Adding 3 to all parts of the inequality:
23 + 3 < 3x - 3 + 3 ≤ -66 + 3
Simplifying:
26 < 3x ≤ -63
Next, divide all parts of the inequality by 3:
26/3 < 3x/3 ≤ -63/3
Simplifying:
8.67 < x ≤ -21
Therefore, the solution to the inequality is:
x ∈ (-∞, -21]
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