(a) For f(x) = 2x^2 for 0 ≤ x ≤ C, integrating and solving for C gives us the value of C. (b) For f(x) = Cx for 0 ≤ x ≤ 3, integrating and solving for C gives us the value of C. (c) For f(x) = ex for 0 ≤ x ≤ C, integrating and solving for C gives us the value of C. (d) Similarly, for the remaining functions (e), (f), (g), and (h), integrating and solving for C will give us the values of C in each case.
In order to find the value of C for each probability density function, we need to ensure that the integral of the function over its given range equals 1, since the total area under the probability density function represents the probability of the random variable occurring.
(a) To find C for f(x) = 2x^2 for 0 ≤ x ≤ C, we need to integrate the function over its given range and set it equal to 1:
∫[0,C] 2x^2 dx = 1
After integrating and solving for C, we can determine the value.
(b) For f(x) = Cx for 0 ≤ x ≤ 3, we integrate the function and set it equal to 1:
∫[0,3] Cx dx = 1
After integrating and solving for C, we can find its value.
(c) For f(x) = ex for 0 ≤ x ≤ C, we integrate the function and set it equal to 1:
∫[0,C] ex dx = 1
After integrating and solving for C, we can determine the value.
(d), (e), (f), (g), and (h) follow a similar process. By integrating each function over its given range and equating the result to 1, we can solve for C and find its value in each case.
By finding the appropriate antiderivatives and solving the resulting equations, we can determine the values of C for each probability density function.
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9. A random variable X is distributed according to X~ N(= 25,0² =9) (a) Determine such M so that P(X < M) = 0.95. (b) Determine the median.
The standard normal distribution has a mean of 0 and a standard deviation of 1. M ≈ 30.935. The median of the distribution is also 25.
(a) To find M, we first need to convert the given values of mean and standard deviation to the standard normal distribution. This can be done by using the formula Z = (X - μ) / σ, where Z is the Z-score, X is the value of interest, μ is the mean, and σ is the standard deviation. In this case, we have X ~ N(25, 9). Substituting the values into the formula, we get Z = (X - 25) / 3. Now we need to find the Z-score that corresponds to the desired probability of 0.95. Using a standard normal distribution table or a calculator, we find that the Z-score corresponding to a cumulative probability of 0.95 is approximately 1.645. Setting Z equal to 1.645, we can solve for X: (X - 25) / 3 = 1.645. Solving for X, we get X ≈ 30.935. Therefore, M ≈ 30.935.
(b) The median is the value that divides the distribution into two equal halves. In a normal distribution, the median is equal to the mean. In this case, the mean is given as 25. Therefore, the median of the distribution is also 25.
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Stay on the same data set: GPA and Weight Form a table using the way the student feels about their weight. Insert the table into two columns into StatCrunch. At the 1% signficance level do the data provide sufficient evidence to conclude that the way students feel about their weight is the same? Run a Goodness of Fit Test. Show all 6 steps: 1. State the null and alternative hypotheses 2. State the significance level 3. State the test statistic 4. State the P-value 5. State the Decision 6. Interpret
1. State the null and alternative hypotheses;Null hypothesis (H0): The distribution of the way students feel about their weight is the same.
Alternative hypothesis (Ha): The distribution of the way students feel about their weight is not the same.
2. State the significance level:
The significance level (α) is given as 1% or 0.01.
3. State the test statistic:
For a Goodness of Fit Test, we typically use the Chi-square (χ²) test statistic.
4. State the P-value:
The P-value represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true. We will obtain the P-value from the Chi-square distribution.
5. State the decision:
We will compare the P-value to the significance level (α). If the P-value is less than or equal to α, we reject the null hypothesis. Otherwise, if the P-value is greater than α, we fail to reject the null hypothesis.
6. Interpret:
Based on the decision, we interpret the results in the context of the study.
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Listed below are the lead concentrations in μg/g measured in different traditional medicines. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 18μg/g. Assume that the sample is a simple random sample. 6.5 14.5 18,5 14.5 18.5 17 2.5 12.5 15 16
A. H0 :μ>18μg/g H1 :μ<18μg/g B. H0 :μ=18μg/g H1 :μ=18μg/g
C. H0 :μ=18μg/g H1 :μ<18μg/g D. H0 :μ=18μg/g H1 :μ>18μg/g Determine the test statistic. (Round to two decimal places as needed.)
the test statistic is approximately -2.125.
To test the claim that the mean lead concentration for all such medicines is less than 18 μg/g, we can use a one-sample t-test. The hypotheses are as follows:
H0: μ ≥ 18 μg/g (Null hypothesis)
H1: μ < 18 μg/g (Alternative hypothesis)
The test statistic for a one-sample t-test is given by:
t = ([tex]\bar{X}[/tex] - μ) / (s / √n)
where [tex]\bar{X}[/tex] is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
Given the data: 6.5, 14.5, 18.5, 14.5, 18.5, 17, 2.5, 12.5, 15, 16
The sample mean ([tex]\bar{X}[/tex]) is calculated as the average of the data:
[tex]\bar{X}[/tex] = (6.5 + 14.5 + 18.5 + 14.5 + 18.5 + 17 + 2.5 + 12.5 + 15 + 16) / 10 = 14.3
The sample standard deviation (s) can be calculated using the formula:
s = √[Σ(xi - [tex]\bar{X}[/tex])² / (n - 1)]
= √[(6.5 - 14.3)² + (14.5 - 14.3)² + (18.5 - 14.3)² + (14.5 - 14.3)² + (18.5 - 14.3)² + (17 - 14.3)² + (2.5 - 14.3)² + (12.5 - 14.3)² + (15 - 14.3)² + (16 - 14.3)² / (10 - 1)]
= √[74.7 / 9]
≈ 3.076
Now, we can calculate the test statistic:
t = ([tex]\bar{X}[/tex] - μ) / (s / √n)
= (14.3 - 18) / (3.076 / √10)
≈ -2.125
Therefore, the test statistic is approximately -2.125.
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A pet store owner is trying to decide whether to discontinue selling specialty clothes for pets. She suspects that only 4% of the customers buy specialty clothes for their pets and thinks that she might be able to replace the clothes with more profitable items. Before making a final decision, she decides to keep track of the total number of customers for a day and whether they purchase specialty clothes. a. The owner had 275 customers that day. Assuming this was a typical day for her store, what would be the mean and standard deviation of the number of customers who buy P a ge L ADM 2303- Spring/Summer 2022 specialty clothes for their pets each day? b. Could we use a normal distribution to approximate the binomial distribution in this case? c. What is the probability of less than 9 customers purchasing specialty clothes for their pets that day? d. What is the probability of more than 18 customers purchasing specialty clothes for their pets that day? ( 2 points) e. Assuming that 18 customers bought specialty clothes on the specified day, the owner thought that her 4% estimate must have been too low. Provide reasoning for agreeing or disagreeing with the statement?
a)The mean of the number of customers specialty clothes is 11 and the standard deviation is approximately 3.23.
b)The binomial distribution are generally satisfied when n * p ≥ 5 and n * (1 - p) ≥ 5.
c)The binomial probability formula with n = 275 and p = 0.04, we can calculate this probability.
d)The binomial probability formula with n = 275 and p = 0.04, we can calculate this probability.
e)A conclusion about the accuracy of the estimate, a larger sample of days or additional data needed.
The mean (μ) and standard deviation (σ) of the number of customers who buy specialty clothes for their pets each day calculated using the properties of the binomial distribution.
The mean is given by the formula: μ = n × p, where n is the total number of customers (275) and p is the probability of buying specialty clothes (0.04).
μ = 275 × 0.04 = 11
The standard deviation is given by the formula: σ = √(n × p × (1 - p))
σ = √(275 × 0.04 × (1 - 0.04)) = √(10.44) ≈ 3.23
A normal distribution to approximate the binomial distribution in this case. The conditions for using the normal approximation to the binomial distribution are generally satisfied when n × p ≥ 5 and n × (1 - p) ≥ 5. In this case, 275 × 0.04 = 11 ≥ 5 and 275 × (1 - 0.04) = 264 ≥ 5, so the conditions are met.
To find the probability of less than 9 customers purchasing specialty clothes for their pets, use the binomial probability formula:
P(X < 9) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 8)
d. To find the probability of more than 18 customers purchasing specialty clothes for their pets, use the complement rule. The probability of more than 18 customers is equal to 1 minus the probability of 18 or fewer customers:
P(X > 18) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 18))
The fact that 18 customers bought specialty clothes on a specific day does not necessarily imply that the 4% estimate was too low. The number of customers specialty clothes from day to day due to random fluctuations.
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Complete parts a and b.
a. Perform each of the following operations.
1. 13°56'26"+8°43'42"
II. 18°17'-4°45'
b. Express the following without decimals.
i. 0.3°
ii. 18.28°
The correct conversion is 18.28° can be expressed as 18°16' without decimals.
a.To perform the operation 13°56'26" + 8°43'42", we add the degrees, minutes, and seconds separately:
Degrees: 13° + 8° = 21°
Minutes: 56' + 43' = 99' = 1°39' (since 60 minutes = 1 degree)
Seconds: 26" + 42" = 68" = 1'8" (since 60 seconds = 1 minute)
Therefore, 13°56'26" + 8°43'42" = 21°1'39" + 1°8" = 22°9'47".
II. To perform the operation 18°17' - 4°45', we subtract the degrees, minutes, and seconds separately:
Degrees: 18° - 4° = 14°
Minutes: 17' - 45' = -28' = -28'
Seconds: There are no seconds in this operation.
Therefore, 18°17' - 4°45' = 14°-28'.
b.i. To express 0.3° without decimals, we convert it to minutes:
0.3° = 0°18'
Therefore, 0.3° can be expressed as 0°18' without decimals.
ii. To express 18.28° without decimals, we split it into degrees and minutes:
18.28° = 18° + 0.28°
Since 1 degree = 60 minutes, we can convert 0.28° to minutes:
0.28° = 0°16.8' = 0°16' + 0.8'
Therefore, 18.28° can be expressed as 18°16' without decimals.
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True or False: A survey of autos parked in student and staff lots at a large college recorded the make, country of origin, type of vehicle (car, SUV, etc.) and age. The make of the auto is used for analysis and is considered categorical
The statement "The make of the auto is used for analysis and is considered categorical" is true, and it is a nominal categorical variable.
The main reason for the statement being true is that the make of an auto is considered a categorical variable because it is in a specific group that cannot be ordered. The make of a car cannot be arranged in any order, but it can be counted. It is divided into groups that contain the same values. Categorical variables have two types: nominal and ordinal, but make is nominal because there is no way to put car makes in any type of order. For example, Toyota cannot be considered greater or less than BMW. Therefore, a survey of autos parked in student and staff lots at a large college recorded the make, country of origin, type of vehicle (car, SUV, etc.) and age. The make of the auto is used for analysis and is considered categorical.
Thus, the statement "The make of the auto is used for analysis and is considered categorical" is true, and it is a nominal categorical variable.
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Given a normal distribution with u = 101 and a=8, and given you select a sample of n = 16, complete (Type an integer or decimal rounded to four decimal places as needed.)
b. What is the probability that X is between 95 and 97.5?
P(95
(Type an integer or decimal rounded to four decimal places as needed.)
c. What is the probability that X is above 101.6?
P(X>101.6)=3821
(Type an integer or decimal rounded to four decimal places as needed.)
Gi
d. There is a 63% chance that X is above what value?
A (Type an integer or decimal rounded to two decimal places as needed.)
Th
The correct answer is b) the probability of z score being between -2 and -1.25 is 0.0778 or 0.078 approximately, i.e., P(95 < X < 97.5) = 0.078c. c) P(X>101.6) = 0.3085d. and d) there is a 63% chance that X is above 101.74 (rounded to two decimal places).
Given a normal distribution with u = 101 and a =8, and given you select a sample of n = 16.
b. What is the probability that X is between 95 and 97.5?
Solution: For X = 95 and z score = (95 – 101) / (8 / √16) = -2
For X = 97.5 and z score = (97.5 – 101) / (8 / √16) = -1.25
We can get the z-scores using the z-table.
Using the z-table, the probability of z score being between -2 and -1.25 is 0.0778 or 0.078 approximately, i.e., P(95 < X < 97.5) = 0.078c.
c) What is the probability that X is above 101.6?
Solution: For X = 101.6 and z score = (101.6 – 101) / (8 / √16) = 0.5
The area under the standard normal distribution curve to the right of z = 0.5 is 0.3085 approximately.
Thus, P(X>101.6) = 0.3085d.
d) There is a 63% chance that X is above what value?
Solution: From the standard normal distribution table, the z score that corresponds to 63% is z = 0.37.
Using this value, we can calculate the corresponding value of X as:0.37 = (X – 101) / (8 / √16)
Solving for X, we get X = 101 + (0.37 × 2) = 101.74
Therefore, there is a 63% chance that X is above 101.74 (rounded to two decimal places).
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About 77% of young adults think they can achieve the American dream. 1.25 pts Determine if the following statement is true or false. The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n = 30. True False
False. The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n = 30.
About 77% of young adults think they can achieve the American dream.
The sample proportion is p = 0.77. And the sample size is n = 40.T
o determine if the following statement is true or false:
The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n = 30.
The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal.
The given statement is False because n = 40, not 30.
Hence, the Central Limit Theorem can be applied to sample proportion.
If the sample size is large enough (n > 30) and the sample satisfies the
np > 10 and nq > 10, where q = 1 - p, then we can use the normal distribution to approximate the sample proportion as shown below:$$\frac{\hat p-p}{\sqrt{\frac{pq}{n}}}\sim N(0,1)$$
Hence, the distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal, as n > 30 and np = 31 > 10, nq = 9 > 10. T
herefore, the given statement is false.
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hich of the following integrals is equal to the line integral where C is the curve parameterized by x=t 3
,y=3t for 0≤t≤1 ? ∫ 0
1
9t 4
+9
dt
∫ 0
1
3t 4
9t 4
+9
dt
∫ 0
1
3t 4
dt
∫ 0
1
27t 6
dt
∫ 0
3
∫ 0
1
xydxdy
∫ C
xyds 1
The correct option is ∫ 0
1
3t^4 / (9t^4 + 9) dt.
The line integral of a function F(x, y) over a curve C parameterized by x = t^3, y = 3t for 0 ≤ t ≤ 1 is given by:
∫C F(x, y) ds
To determine which of the given integrals is equal to this line integral, we need to express ds in terms of the parameter t and find the appropriate form.
The differential ds for a curve parameterized by x = x(t), y = y(t) is given by:
ds = √(dx^2 + dy^2) = √((dx/dt)^2 + (dy/dt)^2) dt
In this case, x = t^3 and y = 3t, so we have:
dx/dt = 3t^2
dy/dt = 3
Substituting these values, we have:
ds = √((3t^2)^2 + (3)^2) dt
= √(9t^4 + 9) dt
= 3√(t^4 + 1) dt
Therefore, the line integral ∫C F(x, y) ds can be written as:
∫C F(x, y) ds = ∫(0 to 1) F(t^3, 3t) * 3√(t^4 + 1) dt
Comparing this expression to the given options, we find that the integral equal to the line integral is:
∫ 0
1
3t^4 / (9t^4 + 9) dt
Thus, the correct option is ∫ 0
1
3t^4 / (9t^4 + 9) dt.
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SAT scores: A college admissions officer sampled 107 entering freshmen and found that 38 of them scored more than 510 on the math SAT. Part: 0/3 Part 1 of 3 (a) Find a point estimate for the proportion of all entering freshmen at this college who scored more than 510 on the math SAT. Round the answer to at least three decimal places. The point estimate for the proportion of all entering freshmen at this college who scored more than 510 on the math SAT IS
The point estimate for the proportion of all entering freshmen at this college who scored more than 510 on the math SAT is approximately 0.355.
To calculate the point estimate, we divide the number of freshmen who scored more than 510 on the math SAT (38) by the total number of freshmen sampled (107). This gives us a proportion of 0.355, which represents the estimated proportion of all entering freshmen at the college who scored above the given threshold.
In other words, based on the sample data, it is estimated that approximately 35.5% of all entering freshmen at this college scored more than 510 on the math SAT. It's important to note that this point estimate is an approximation and may differ from the actual proportion in the entire population of freshmen. However, it provides a useful estimate based on the available sample data.
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test statisfie and select if from the options belone A test atatistic an 0 B. test atatistie =70.341 C. lest statisice =65.354 D. teot stanisic = 2,353 Rifer to the scruatio from Question #1. Finally, decide an a conclusica with cociect contest from the options belors A. We repert the null hypothesis that Trump can win Virginia because the p-value associased with the tent stabstic will clearly be zero. Blden will win the state. B. We reject the nuil hypobhesis that Trump can win Colorado because the p-value associated with the test statistic will clearfy be zero. Bicen wal win the state. C. Wo fail to reject the null hypothesis that Trump can win Colorado because our fest statistic is not below the value of alpha. D. We fail to reject the null hypothesis that
The correct answer for the conclusion is B. We reject the null hypothesis that Trump can win Colorado because the p-value associated with the test statistic will clearly be zero.
How to explain the informationThe p-value associated with a test statistic of 0 will be zero, because the probability of observing a test statistic of exactly 0 is zero. This means that the null hypothesis can be rejected with certainty, and we can conclude that the alternative hypothesis is true.
In the context of the scenario, this means that we can reject the null hypothesis that Trump can win Colorado, and conclude that Biden will win the state.
The correct answer for the conclusion is B. We reject the null hypothesis that Trump can win Colorado because the p-value associated with the test statistic will clearly be zero. Biden will win the state.
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If the current in a circuit is 2-j5 volts and the resistance is 1+j3 ohms, what is the voltage? a.) 13+j11 amps b.) 17+j11 amps c.) 17+jamps d.) 13+j amps in SUBMIT MY ANSWER Report an issue with this question O fie ex
The voltage in the circuit can be calculated by multiplying the current and the resistance. Given a current of 2-j5 volts and a resistance of 1+j3 ohms, the voltage is 17+j11 amps.
To calculate the voltage, we can use Ohm's Law, which states that voltage (V) is equal to the product of current (I) and resistance (R). In this case, we have V = I * R, where I = 2-j5 volts and R = 1+j3 ohms. Multiplying these values, we get V = (2-j5) * (1+j3). Using the distributive property, we expand the expression to V = 2 + 6j - j5 -j². Simplifying further, we combine like terms and substitute j² with -1 (since j² is equal to -1). Thus, V = 2 - 5j - 1 + 6j = 1 + j. Therefore, the voltage in the circuit is 1+j amps.
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Two randomly selecied puestions (chosen by Professo Dicken-Conlin) will be grated for erodit All graded guestions are werth 10 points, Uve nurs puges if acceaury, but make naie you ciearly indicaie where fie fest of your anwase in. 1. The demand for a pair of mitoncy cle riding flever is summarized in the squation Q
b
=900−30P, where price is in dellars per peir. a. Suppose the price of a pair of gloves is 310 : 1. What is the gaasaiify demanded? 4. Show the deranand curve gerphically, Notice that you arr given the DEMAND curve, a0 the invere demand carve, so you may want so tewrie it as the inverse DEMAND earve, whict is ia sicpe intercept form. ii. Numerically and groplically calculate the consiamer iarphus and interyre the namber you calculate (ase imuation). b. Suppose a government rupelesice leads poodocen to lacrease the poice from 510 w 120 1. What is the quantity demanded? if. Laituitively will consamer surplus increase or decrease if the price increasest Explain iif. Calculate mumerically the change in conamer maplus, In other worthe, by how mach does it iacrease or dectear? W. Show the new consumar surplus graphically.
Quantity demanded at a price of $310: 840 pairs. Graph the demand curve accordingly. Consumer surplus: $21,000. Price increase to $120 decreases quantity demanded to 780 pairs, reducing consumer surplus by $3,600.
When the price of a pair of gloves is $310, we can substitute this value into the demand equation Qd = 900 - 30P to find the quantity demanded.
To calculate the consumer surplus, we need to find the area between the demand curve and the price line, both numerically and graphically.
The numerical calculation involves integrating the area under the demand curve and above the price line. The interpretation of the consumer surplus value obtained will indicate the net benefit to consumers.
With the price change from $510 to $120, we can substitute the new price into the demand equation to find the corresponding quantity demanded. Intuitively, when the price increases, the consumer surplus is expected to decrease due to a higher cost for consumers.
Numerically, the change in consumer surplus can be calculated by comparing the consumer surplus before and after the price change.
Note: The specific calculations, interpretations, and graphical representations depend on the methods and tools used in analysis.
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Don't Be Late!
1. In a study of proctored and non-proctored math tests, researchers obtained the data below. Use a 0.05 significance level that the students taking the proctored tests get a lower mean grade than the students taking non-proctored tests.
Group 1 (proctored):
n=30, x = 75.72, s = 11.64
Group 2 (non-proctored):
n=32, x=87.51, s = 20.97
The students taking proctored tests have a lower mean grade than the students taking non-proctored tests with a significance level of values 0.05.
To determine if there is a significant difference between the mean grades of students taking proctored tests and non-proctored tests perform a two-sample t-test.
Null hypothesis (H0): The mean grade of students taking proctored tests is equal to or greater than the mean grade of students taking non-proctored tests.
Alternative hypothesis (Ha): The mean grade of students taking proctored tests is lower than the mean grade of students taking non-proctored tests.
Group 1 (proctored):
n1 = 30, x1 = 75.72, s1 = 11.64
Group 2 (non-proctored):
n2 = 32, x2 = 87.51, s2 = 20.97
calculate the test statistic (t) using the formula:
t = (x1 - x2) / √((s1² / n1) + (s2² / n2))
Substituting the values:
t = (75.72 - 87.51) / √((11.64² / 30) + (20.97² / 32))
Calculating this value t =-2.356
To determine if this test statistic is significant at a significance level of 0.05, it with the critical value from the t-distribution table with degrees of freedom (df) given by:
df = (s1² / n1 + s2² / n2)² / [((s1² / n1)² / (n1 - 1)) + ((s2² / n2)² / (n2 - 1))]
Substituting the values:
df =59.03
The critical value for a one-tailed t-test at a significance level of 0.05 and degrees of freedom (df) =59.03 is approximately -1.671.
Since the test statistic t = -2.356 is smaller than the critical value -1.671, the null hypothesis.
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Karissa begins to solve the equation StartFraction one-half EndFraction left-parenthesis x minus 14 right-parenthesis plus 11 equals StartFraction one-half EndFraction x minus left-parenthesis x minus 4 right-parenthesis.. Her work is correct and is shown below.
Three lines of math. The first line, StartFraction one-half EndFraction left-parenthesis x minus 14 right-parenthesis plus 11 equals StartFraction one-half EndFraction x minus left-parenthesis x minus 4 right-parenthesis. The second line, StartFraction one-half EndFraction x minus 7 plus 11 equals StartFraction one-half EndFraction x minus x plus 4. The third line StartFraction one-half EndFraction x plus 4 equals negative StartFraction one-half EndFraction x plus 4.
StartFraction one-half EndFraction x minus 7 plus 11 equals StartFraction one-half Endfraction x minus x plus 4.
StartFraction one-half EndFraction x plus 4 equals negative StartFraction one-half Endfraction x plus 4.
When she subtracts 4 from both sides, Startfraction one-half EndFraction x equals negative StartFraction one-half EndFraction x. results. What is the value of ?
The equation Start Fraction one-half End Fraction x - 7 + 11 = Start Fraction one-half End Fraction x - x + 4 leads to the conclusion that the value of x can be any real number.
Based on the given information, let's analyze the steps taken by Karissa and determine the value of x.
We start with the equation:
Start Fraction one-half End Fraction left-parenthesis x - 14 right-parenthesis + 11 = Start Fraction one-half End Fraction x - left-parenthesis x - 4 right-parenthesis.
Karissa's first step is to distribute the fractions on both sides of the equation:
Start Fraction one-half End Fraction x - 7 + 11 = Start Fraction one-half End Fraction x - x + 4.
Simplifying further, we combine like terms:
Start Fraction one-half End Fraction x + 4 = Start Fraction one-half End Fraction -x + 4.
The next step is to subtract x from both sides of the equation:
Start Fraction one-half End Fraction x + 4 - x = Start Fraction one-half End Fraction -x + 4 - x.
Simplifying gives us:
Start Fraction one-half End Fraction x - x + 4 = Start Fraction one-half End Fraction -2x + 4.
Now, let's subtract 4 from both sides of the equation:
Start Fraction one-half End Fraction x - x = Start Fraction one-half End Fraction -2x.
Simplifying further:
Start Fraction one-half End Fraction x = - Start Fraction one-half End Fraction x.
From this step, we can observe that the variable x cancels out on both sides of the equation.
This means that no matter what value we assign to x, the equation remains true.
Therefore, the value of x can be any real number.
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Define a relation ∼ on Z by a∼b if a ≤ b (e.g 4∼5, since 4≤5, while 7≁5). (i) Is ∼ reflexive? (ii) Is ∼ symmetric? (iii) Is ∼ transitive?
Let us consider the relation ∼ on Z by a ∼ b if a ≤ b and the terms reflexive, symmetric, and transitive for each property.
(i) Reflexive: A relation ~ on a set Q is called reflexive if every element of Q is related to itself. That is, for all a ∈ Q, a ~ a. In this case, the relation ∼ on Z is reflexive. a ∼ a for any a ∈ Z.
(ii) Symmetric: A relation ~ on a set P is called symmetric if for all a, b ∈ P, if a ~ b, then b ~ a. In this case, the relation ∼ on Z is not symmetric. For example, 8 ∼ 9 but 9 is not ∼ 8.
(iii) Transitive: A relation ~ on a set S is called transitive if for all a, b, c ∈ S, if a ~ b and b ~ c, then a ~ c. In this case, the relation ∼ on Z is transitive. If a ≤ b and b ≤ c, then a ≤ c. So, a ∼ b and b ∼ c implies a ∼ c.
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Find the probability by using Empirical Rule for the following (Do not use Z-table); Given population mean of μ =7 and a standard deviation of σ = 2,
Find the probability of P(x>7).
According to the empirical rule, Probability = approximately 99.7% .
Approximately 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and around 99.7% falls within three standard deviations.
The empirical rule, also known as the 68-95-99.7 rule, provides a way to estimate probabilities based on the standard deviation of a population. Given a population mean (μ) of 7 and a standard deviation (σ) of 2, we can use the empirical rule to find the probabilities for different ranges of values. According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and around 99.7% falls within three standard deviations.
Using the empirical rule, we can estimate the probabilities for different ranges of values based on the given mean (μ) and standard deviation (σ).
Within one standard deviation of the mean:
The range is from μ - σ to μ + σ.
Probability = approximately 68%
Within two standard deviations of the mean:
The range is from μ - 2σ to μ + 2σ.
Probability = approximately 95%
Within three standard deviations of the mean:
The range is from μ - 3σ to μ + 3σ.
Probability = approximately 99.7%
For the given population mean of μ = 7 and a standard deviation of σ = 2, we can use the empirical rule to estimate the probabilities as described above. These probabilities provide a rough estimate of how likely it is for a randomly selected data point to fall within each respective range. Keep in mind that the empirical rule assumes a normal distribution and may not be precise for all data sets.
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An internship student in Public Bank in Malaysia was asked by the bank manager to find out the mean amount of mortgage paid per month by landlords in area Y. He randomly selected a sample of 125 landlords from this area. Based on the data collected, the results revealed that the bank received a pay of an average of RM1798 per month for their mortgages. Suppose that the monthly mortgage payments of all landlords have a normal distribution and it is known that the population variance of such mortgages is RM53824. (a) Determine (C1) the point estimate of μ. (1 mark) (b) Construct (C3) a 95\% confidence interval for the mean amount of mortgage paid per month by all landlords in area Y. Give your answer in three decimal places. (c) Suppose the confidence interval obtained in part (b) is too wide. Suggest (C3) all possible alternatives to reduce the width of this interval. Give (C1) your opinion and reasoning on which alternative is the best.
a. The point estimate of μ (population mean) is RM1798.
b. The 95% confidence interval for the mean amount of mortgage paid per month by all landlords in area Y is approximately (RM1757.33, RM1848.67).
c. Alternatives to reduce the width of the confidence interval include increasing the sample size, decreasing the confidence level, or reducing the population standard deviation. Increasing the sample size is the best option to obtain a narrower interval.
How to find the point estimate of μ (population mean) is the average mortgage payment from the sample?(a) The point estimate of μ (population mean) is the average mortgage payment from the sample, which is RM1798 per month.
How to construct a 95% confidence interval for the mean amount of mortgage paid per month by all landlords in area Y?(b) To construct a 95% confidence interval for the mean amount of mortgage paid per month by all landlords in area Y, we'll use the formula:
CI = [tex]\bar{X}[/tex]± z * (σ / √n)
Where:
[tex]\bar{X}\\[/tex] is the sample mean (point estimate) = RM1798
z is the z-score corresponding to the desired confidence level of 95% (z = 1.96 for a 95% confidence level)
σ is the population standard deviation = √53824 ≈ 231.99
n is the sample size = 125
Plugging in these values, we can calculate the confidence interval:
CI = 1798 ± 1.96 * (231.99 / √125)
Calculating this expression:
CI ≈ 1798 ± 1.96 * (231.99 / 11.18)
CI ≈ 1798 ± 1.96 * 20.76
CI ≈ 1798 ± 40.67
The 95% confidence interval for the mean amount of mortgage paid per month by all landlords in area Y is approximately (1757.33, 1848.67) in RM.
What are all possible alternatives to reduce the width of this interval?(c) To reduce the width of the confidence interval, we can consider the following alternatives:
Increase the sample size:By increasing the sample size, we reduce the standard error and thus decrease the width of the confidence interval. Collecting data from more landlords would provide more precise estimates of the population mean.
Decrease the desired confidence level:If a narrower confidence interval is acceptable, we can choose a lower confidence level. However, this comes at the cost of reduced certainty about the true population mean.
Reduce the population standard deviation:If it is possible to decrease the variability in mortgage payments among landlords in area Y, the confidence interval will become narrower. However, this may not be within the control of the intern.
Among these alternatives, the best option would be to increase the sample size.
By collecting data from a larger number of landlords, the sample mean becomes more representative of the population mean, resulting in a narrower confidence interval.
This would provide a more precise estimate of the mean amount of mortgage paid per month by all landlords in area Y.
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A population of values has a normal distribution with = 150.4 and = 70. a. Find the probability that a single randomly selected value is between 148.6 and 155.2.
The probability that a single randomly selected value is between 148.6 and 155.2 is approximately 0.0394, or 3.94%.
To find the probability that a single randomly selected value is between 148.6 and 155.2 in a normal distribution with a mean (μ) of 150.4 and a standard deviation (σ) of 70, we can use the standard normal distribution.
First, we need to standardize the values of 148.6 and 155.2 using the formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
For 148.6:
z = (148.6 - 150.4) / 70 = -0.026
For 155.2:
z = (155.2 - 150.4) / 70 = 0.068
Next, we can use a standard normal distribution table or a calculator to find the probabilities associated with these z-values.
Using the standard normal distribution table, we can find the cumulative probabilities for these z-values. The cumulative probability for -0.026 is approximately 0.4893, and the cumulative probability for 0.068 is approximately 0.5287.
To find the probability that a single randomly selected value is between 148.6 and 155.2, we subtract the lower probability from the higher probability:
P(148.6 ≤ X ≤ 155.2) = P(X ≤ 155.2) - P(X ≤ 148.6)
= 0.5287 - 0.4893
= 0.0394
Therefore, the probability that a single randomly selected value is between 148.6 and 155.2 is approximately 0.0394, or 3.94%.
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2. Twenty beluga whales were randomly chosen from the MacKenzie Delta. For cach whale, liver selenium (X, with units ag/g dry weight) and tooth selenium (Y, with units were measured. The results are summarized as follows y=156.7 r-0.5273, SS(residual) 17573.3 a. (5 pts) Calculate the linear regression of Y on X. (find and by in y- + ng g dry weight) = 22.68, S 12.54, s38.04
The linear regression equation of Y on X is Y = 22.68 - 0.5273X. The intercept, denoted as "a," is 22.68, and the slope, denoted as "b," is -0.5273. Additionally, the standard error of the intercept, denoted as "sᵃ," is 12.54, and the standard error of the slope, denoted as "sᵇ," is 38.04.
The linear regression equation represents the best-fitting line that describes the relationship between the two variables, Y and X. In this case, the equation suggests that as X (liver selenium) increases, Y (tooth selenium) decreases. The intercept of 22.68 indicates the expected value of Y when X is zero, and the negative slope of -0.5273 implies that, on average, for each unit increase in X, Y decreases by 0.5273 units.
The standard errors, sᵃ and sᵇ, provide information about the precision of the estimated intercept and slope, respectively. These values help assess the uncertainty associated with the regression coefficients. A smaller standard error indicates a more precise estimate. In this case, the standard errors are 12.54 for the intercept and 38.04 for the slope.
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Brody is driving on a long road trip. He currently has 9 gallons of gas in his car. Each hour that he drives, his car uses up 2 gallons of gas. How much gas would be in the tank after driving for 2 hours? How much gas would be left after � t hours? Gas left after 2 hours: Gas left after � t hours:
Brody has 5 gallons of gas left in his car after driving for 2 hours
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
Brody has 9 gallons of gas and each hour, it reduces by 2 gallons of gas. Let us assume that he drives for t hours.
If y represent the amount of gas remaining after time t, then:
y = 9 - 2t
Let us assume he is driving for 2 hours, therefore:
y = 9 - 2(2)
y = 5
He has 5 gallons of gas left
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Independent samples t-test by hand 1. Let’s say we have two groups, group 1 = a sample of athletes & group 2 = a sample of non-athletes, who are asked about the number
of hours they exercise per day. Group 1 had a mean of 1 = 4.5 and Group 2 had a mean of 2 = 1.7. Sample size for each group was N1 = 9 and N2 = 9. Standard deviations for group 1 and group 2 are s1 = .9 and s2 = 1.3. We want to know if the sample means differ from one another and decide to do an independent-samples t test. Please compute the observed t statistic by hand. Report the t statistic using three decimal places. For full credit, be sure to show all of your work.
The observed t statistic is approximately -7.406.
The observed t statistic is calculated to determine whether the means of two independent groups, athletes and non-athletes, significantly differ from each other in terms of the number of hours they exercise per day. In this scenario, Group 1 (athletes) had a mean of 4.5 hours with a standard deviation of 0.9, while Group 2 (non-athletes) had a mean of 1.7 hours with a standard deviation of 1.3. Both groups consisted of 9 participants.
To calculate the observed t statistic, we use the formula:
t = (mean1 - mean2) / √((s1² / N₁) + (s2² / N₂))
Plugging in the given values, we have:
t = (4.5 - 1.7) / √((0.9² / 9) + (1.3² / 9))
t = 2.8 / √(0.01 + 0.0151)
t = 2.8 / √(0.0251)
t = 2.8 / 0.1584
t ≈ -7.406
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Evaluate the following double integral. Every antiderivative must be found using an integration technique. 7/42 cos 0 rr dr do 0 sece (12pts) Evaluate the following double integral. Every antiderivative must be found using an integration technique. 7/42 cos 0 rr dr do 0 sece
Since cos(0) = 1, the integral becomes ∫∫(7/42) dxdy. The given double integral ∫∫(7/42)cos(0) dxdy simplifies to ∫∫(7/42) dxdy. Evaluating this integral results in the value of (7/42) times the area of the region of integration.
1. The integral of a constant with respect to x yields the product of the constant and the variable of integration, in this case, x. Therefore, integrating (7/42) with respect to x gives us (7/42)x + C1, where C1 is the constant of integration.
2. Next, we integrate (7/42)x + C1 with respect to y. The limits of integration for y are 0 to sec(e). Integrating (7/42)x + C1 with respect to y, we get (7/42)x*y + C1*y + C2, where C2 is the constant of integration with respect to y.
3. Now, we evaluate the double integral by substituting the limits of integration. For y, we have 0 to sec(e), and for x, we have 0 to r.
(7/42) times the double integral ∫∫dxdy becomes (7/42) times the integral of (7/42)x*y + C1*y + C2 with respect to y, evaluated from 0 to sec(e).
4. Plugging in the limits of integration, we have (7/42)[(7/42)r*sec(e) + C1*sec(e) + C2 - (7/42)(0) - C1(0) - C2]
Simplifying, the result is (7/42)[(7/42)r*sec(e) + C1*sec(e) + C2].
5. Thus, the value of the double integral ∫∫(7/42)cos(0) dxdy is (7/42)[(7/42)r*sec(e) + C1*sec(e) + C2], which is (7/42) times the area of the region of integration, adjusted by the constants of integration.
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The lifetime of lightbulbs that are advertised to last for 5900 hours are normally distributed with a mean of 6165.5 hours and a standard deviation of 150 hours. What is the probability that a bulb lasts longer than the advertised figure? Probability =
The lifetime of light bulbs that are advertised to last for 5900 hours are normally distributed with a mean of 6165.5 hours and a standard deviation of 150 hours.
To find the probability that a bulb lasts longer than the advertised figure, we need to calculate the z-score of 5900. Then, we will use the z-score table to find the probability of the bulb lasting longer than 5900 hours.
z-score formula is given by: Z = (X - μ) / σ, where, X = 5900 hours μ = 6165.5 hours σ = 150 hours
Plugging these values in the formula, we get :Z = (5900 - 6165.5) / 150
Z = -0.177
Let us check the z-table to find the probability for z = -0.177 from the standard normal distribution table, the area to the left of the z-score -0.177 is 0.4306.
Since we want to find the probability that a bulb lasts longer than 5900 hours, we need to subtract the value obtained from 1. Thus, the probability that a bulb lasts longer than the advertised figure is: 1 - 0.4306 = 0.5694, which is approximately equal to 0.57 or 57%.
Therefore, the probability that a bulb lasts longer than the advertised figure is 0.57 or 57%.
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Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 52 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean μ=52 tons and standard deviation σ=1.5 ton.
What is the probability that one car chosen at random will have less than 51.5 tons of coal? (Round your answer to four decimal places.)
The probability that one car chosen at random will have less than 51.5 tons of coal is approximately 0.3707, rounded to four decimal places.
To find the probability that one randomly chosen car will have less than 51.5 tons of coal, we can use the normal distribution and the given mean (μ = 52 tons) and standard deviation (σ = 1.5 tons).
First, we need to calculate the z-score for the value 51.5 tons using the formula:
z = (x - μ) / σ
Substituting the given values:
z = (51.5 - 52) / 1.5 = -0.3333
Next, we can use a standard normal distribution table or a calculator to find the cumulative probability associated with the z-score of -0.3333. The cumulative probability represents the area under the standard normal distribution curve to the left of the given z-score.
Looking up the z-score of -0.3333 in the standard normal distribution table, we find that the cumulative probability is 0.3707.
Therefore, the probability that one car chosen at random will have less than 51.5 tons of coal is approximately 0.3707, rounded to four decimal places.
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Suppose that you used technology to find the least-squares regression line from a table of values for two variables and the results were displayed as follows. m=4.1136b=18.4717r 2
=0.5947r=0.6841 What can we say about the relationship between the two variables? Be specific.
The displayed results provide information about the estimated slope, y-intercept, and the goodness of fit of the regression line based on the given data.
m = 4.1136: This represents the slope of the regression line. It indicates the change in the dependent variable (y) for every one-unit increase in the independent variable (x). In this case, for each unit increase in x, y is expected to increase by approximately 4.1136 units.
b = 18.4717: This represents the y-intercept of the regression line. It is the value of y when x is equal to zero. In this case, when x is zero, the predicted value of y is approximately 18.4717.
r^2: This is the coefficient of determination, which measures the goodness of fit of the regression line. It represents the proportion of the variance in the dependent variable (y) that can be explained by the independent variable (x). A value between 0 and 1 is typically provided, indicating the strength of the relationship. In this case, r^2 is given but not specified. However, a higher value of r^2 indicates a better fit of the regression line to the data.
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A manufacturer of industrial light bulbs claims that the mean length of life of its light bulbs is 8200 hours. A sample of 20 light bulbs was tested and produced a mean of 8210 hours and variance of 186.69. Assume the population distribution is normally distributed. a. b. Construct a 99% confidence interval for the actual mean length of life of the light bulbs and interpret the interval. Based on the interval in (ii), can you conclude the manufacturer's claim at 99% confidence level? Explain your answer.
The manufacturer's claim at 99% confidence level.
We have the following information:Sample size, n = 20Sample mean, $\bar{x}$ = 8210,
Population variance,\sigma^{2} = 186.69.
We need to calculate the 99% confidence interval for the actual mean length of life of the light bulbs.
The formula for the confidence interval is:$$\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \lt \mu \lt \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$where$\bar{x}$ is the sample mean,$\sigma$ is the population standard deviation, $n$ is the sample size,is the population mean, and$z_{\alpha/2}$ is the critical value of the standard normal distribution at the level of significance, alpha.
The value of $z_{\alpha/2}$ can be found from the standard normal distribution table. Here, we are constructing a 99% confidence interval.
Therefore, the level of significance is \alpha = 0.01$.Now, from the standard normal distribution table, the value of $z_{\alpha/2}$ is 2.58 (approx).
Therefore, the 99% confidence interval for the actual mean length of life of the light bulbs is given as:$$8210 - 2.58\frac{\sqrt{186.69}}{\sqrt{20}} \lt \mu \lt 8210 + 2.58\frac{\sqrt{186.69}}{\sqrt{20}}$$Solving the above inequality, we get:$$8067.53 \lt \mu \lt 8352.4.
Therefore, the 99% confidence interval for the actual mean length of life of the light bulbs is (8067.53, 8352.47).
The manufacturer of industrial light bulbs claims that the mean length of life of its light bulbs is 8200 hours. From the 99% confidence interval obtained in (a), we see that the actual mean length of life of the light bulbs may lie between 8067.53 and 8352.47 hours.
Since 8200 hours lies within this interval, we cannot conclude that the manufacturer's claim is false at 99% confidence level. Therefore, we can say that the sample provides evidence that supports the manufacturer's claim at 99% confidence level.
Hence, the main answer is No, we cannot conclude the manufacturer's claim at 99% confidence level.Explanation:To construct the 99% confidence interval, we use the formula:$$\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \lt \mu \lt \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}Where, $\alpha$ is the level of significance, $z_{\alpha/2}$ is the critical value of the standard normal distribution, $\sigma$ is the population standard deviation, $n$ is the sample size, $\bar{x}$ is the sample mean, and $\mu$ is the population mean.
The calculation for the confidence interval was shown in part (a) as:$$8210 - 2.58\frac{\sqrt{186.69}}{\sqrt{20}} \lt \mu \lt 8210 + 2.58\frac{\sqrt{186.69}}{\sqrt{20}}.
Solving the above inequality, we get:$$8067.53 \lt \mu \lt 8352.47.Therefore, the 99% confidence interval for the actual mean length of life of the light bulbs is (8067.53, 8352.47).
To test whether the manufacturer's claim is true or false, we compare the confidence interval with the given claim. We observe that the claim lies within the confidence interval.
Therefore, we cannot reject the claim at 99% confidence level. Hence, the conclusion is that we cannot conclude the manufacturer's claim at 99% confidence level.
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Freddie scored 19 out of 25 for his maths homework.
Elizabeth scored 23 out of 28 for her maths homework.
a) By finding Freddie and Elizabeth's scores as percentages, work out who had the higher percentage mark.
b) What was the difference in percentage marks between their scores? Give your answer to 1 decimal place.
a) Elizabeth's percentage mark ≈ 82.1% b) Elizabeth had a higher percentage mark than Freddie, and the difference in their percentage marks is approximately 6.1%.
How to calculate Elizabeth's percentage markTo find Freddie and Elizabeth's scores as percentages, we need to divide their scores by the total possible scores and multiply by 100.
a) Freddie's percentage mark:
Freddie's score = 19
Total possible score = 25
Freddie's percentage mark = (19/25) * 100 ≈ 76%
Elizabeth's percentage mark:
Elizabeth's score = 23
Total possible score = 28
Elizabeth's percentage mark = (23/28) * 100 ≈ 82.1%
b) To find the difference in percentage marks between their scores, we subtract Freddie's percentage mark from Elizabeth's percentage mark.
Difference in percentage marks = Elizabeth's percentage mark - Freddie's percentage mark
Difference in percentage marks = 82.1% - 76% ≈ 6.1%
Therefore, Elizabeth had a higher percentage mark than Freddie, and the difference in their percentage marks is approximately 6.1%.
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A student believes that the average grade on the statistics final examination was 87 . A sample of 36 past final examinations was taken. The average grade in the sample was 83.96 with a standard deviation of 12. a. State the null and alternative hypotheses. b. Using the critical value approach, test the hypotheses at the 5% level of significance. Critical Value = Hypothesis Test Conclusion: c. Using the confidence interval approach, test the hypotheses at the 5% level of significance. Confidence Interval: to Hypothesis Test Conclusion:
a) H₀ = 87, H₁ ≠ 87
b) We reject the null hypothesis and have evidence to suggest that the average grade on the statistics final examination is different from 87.
c) We fail to reject the null hypothesis and do not have sufficient evidence to suggest that the average grade on the statistics final examination is different from 87.
a. The null hypothesis (H₀): The average grade on the statistics final examination is 87.
The alternative hypothesis (H₁): The average grade on the statistics final examination is not 87.
b. To test the hypotheses using the critical value approach, we need to calculate the test statistic and compare it to the critical value. The test statistic (t-score) is calculated as:
t = (sample mean - population mean) / (sample standard deviation / √(sample size))
= (83.96 - 87) / (12 / √(36))
= -3.04
Next, we need to determine the critical value for a two-tailed test with a significance level of 5%. Since the sample size is 36, we have degrees of freedom (df) equal to n - 1 = 35. Consulting the t-distribution table or using statistical software, we find the critical value for a two-tailed test with df = 35 and alpha = 0.05 is approximately ±2.032.
Since the absolute value of the test statistic (-3.04) is greater than the critical value (2.032), we reject the null hypothesis. The test result is statistically significant at the 5% level of significance.
Hypothesis Test Conclusion: We reject the null hypothesis and have evidence to suggest that the average grade on the statistics final examination is different from 87.
c. To test the hypotheses using the confidence interval approach, we need to calculate the confidence interval and check if the hypothesized value (87) falls within the interval. The confidence interval is calculated as:
CI = sample mean ± (critical value × (sample standard deviation / √(sample size)))
= 83.96 ± (2.032 × (12 / √(36)))
= 83.96 ± 4.86
The confidence interval is (79.10, 88.82).
Since the hypothesized value of 87 falls within the confidence interval, we fail to reject the null hypothesis. The test result is not statistically significant at the 5% level of significance.
Hypothesis Test Conclusion: We fail to reject the null hypothesis and do not have sufficient evidence to suggest that the average grade on the statistics final examination is different from 87.
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The combined electrical resistance R of two resistors R 1
and R 2
, connected in parallel, is given by the equation below, where R, R 1
, and R 2
are measured in ohms. R 1
and R 2
are increasing at rates of 0.6 and 1.6 ohms per second, respectively. R
1
= R 1
1
+ R 2
1
At what rate is R changing when R 1
=55 ohms and R 2
=72 ohms? (Round your answer to three decimal places.) ohm/sec
The rate at which R is changing when R1=55 ohms and R2=72 ohms is −0.086 ohm/sec.
The given equation is: R1= R1 + R2.
To find the rate at which R is changing, differentiate both sides of the equation with respect to time:
dR1/dt = d(R1+R2)/dt = dR/dt
Given, R1=55 ohms and R2=72 ohms
Then, R = R1R2/(R1+R2)
On substituting the given values, we get R = 29.0196 ohms
Now, dR1/dt = 0.6 ohms/sec and dR2/dt = 1.6 ohms/sec
Using the quotient rule of differentiation, we get:
dR/dt = (R2dR1/dt − R1dR2/dt)/(R1+R2)²
On substituting the given values, we get:
dR/dt = (72×0.6−55×1.6)/(55+72)² ≈ −0.086 ohm/sec
Thus, when R1 = 55 ohms and R2 = 72 ohms, the rate at which R is changing is approximately −0.086 ohm/sec.
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