Question

A retired statistics professor has recorded results for decades. The mean for the score for the...

A retired statistics professor has recorded results for decades. The mean for the score for the population of her students is 82.4 with a standard deviation of 6.5 . In the last year, her standard deviation seems to have changed. She bases this on a random sample of 25 students whose scores had a mean of 80 with a standard deviation of 4.2 . Test the professor's claim that the current standard deviation is different from 6.5 . Use α = .05 .

27) Express the claim in symbolic form.

Express the claim in symbolic form.

Group of answer choices

A )σ > 6.5

B) σ = 6.5

C) σ < 6.5

D) σ ≠ 6.5

E) σ ≥ 6.5

F) σ ≤ 6.5

28) What is the alternative hypothesis, H1?

Group of answer choices

A) σ < 6.5

B) σ ≤ 6.5

C) σ ≠ 6.5

D) σ = 6.5

E) σ > 6.5

F) σ ≥ 6.5

29) Find the critical value(s). (Round to the nearest thousandth. If more than one value is found, enter the smallest critical value.)

30) Find the value of the test statistic. (Round to the nearest ten-thousandth.)

31) What is the statistical conclusion?

Group of answer choices

A)Fail to reject H0

B) Reject H0

32) State the conclusion in words.

Group of answer choices

A) There is not sufficient sample evidence to support the claim that the current standard deviation is different from 6.5 .

B) The sample data support the claim that the current standard deviation is different from 6.5 .

C) There is not sufficient evidence to warrant rejection of the claim that the current standard deviation is different from 6.5 .

D) There is sufficient evidence to warrant rejection of the claim that the current standard deviation is different from 6.5 .

Homework Answers

Answer #1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: σ = 6.5
Alternative Hypothesis, Ha: σ ≠ 6.5

Rejection Region
This is two tailed test, for α = 0.05 and df = 24
Critical value of Χ^2 are 12.401 and 39.364
Hence reject H0 if Χ^2 < 12.401 or Χ^2 > 39.364
critical value = 12.401

Test statistic,
Χ^2 = (n-1)*s^2/σ^2
Χ^2 = (25 - 1)*4.2^2/6.5^2
Χ^2 = 10.02

Fail to rej1ct H0

There is not sufficient sample evidence to support the claim that the current standard deviation is different from 6.5 .

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to...
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach. , , n = 18, H0: μ = 10, Ha: μ < 10, α = 0.01 Group of answer choices Test statistic: t = -4.43. Critical value: t = -2.33. Reject H0. There is sufficient evidence to support the claim that the...
Use a α=0.05 significance level to test the claim that σ=19 if the sample statistics include...
Use a α=0.05 significance level to test the claim that σ=19 if the sample statistics include ,n=10 ,x¯=93, and s=26. The test statistic is The smaller critical number is The bigger critical number is What is your conclusion? A. There is not sufficient evidence to warrant the rejection of the claim that the population standard deviation is equal to 19 B. There is sufficient evidence to warrant the rejection of the claim that the population standard deviation is equal to...
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to...
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach. , , n = 11, H0: μ = 18.7, Ha: μ ≠ 18.7, α = 0.05 Group of answer choices Test statistic: t = 1.03. Critical values: t = ±2.201. Do not reject H0. There is not sufficient evidence to conclude that...
You wish to test the claim that the average IQ score is less than 100 at...
You wish to test the claim that the average IQ score is less than 100 at the .005 significance level. You determine the hypotheses are: Ho: μ=100 H1:μ<100 You take a simple random sample of 76 individuals and find the mean IQ score is 95.5, with a standard deviation of 15.1. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with assuming that it is known. Round to three decimal...
You wish to test the claim that the average IQ score is less than 100 at...
You wish to test the claim that the average IQ score is less than 100 at the .01 significance level. You determine the hypotheses are: H o : μ = 100 H 1 : μ < 100 You take a simple random sample of 60 individuals and find the mean IQ score is 98.7, with a standard deviation of 14.6. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with...
You wish to test the claim that the average IQ score is less than 100 at...
You wish to test the claim that the average IQ score is less than 100 at the .005 significance level. You determine the hypotheses are: H o : μ = 100 H 1 : μ < 100 You take a simple random sample of 95 individuals and find the mean IQ score is 95.2, with a standard deviation of 14.4. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with...
Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for...
Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.1 significance level. a) Identify the correct alternative hypothesis: p=21.21p=21.21 μ>21.21μ>21.21 μ=21.21μ=21.21 μ<21.21μ<21.21 p<21.21p<21.21 p>21.21p>21.21 Give all answers correct to 3 decimal places. b) The test statistic value is:      c) Using the Traditional method, the critical...
Given the sample mean = 22.325, sample standard deviation = 5.8239, and N = 40 for...
Given the sample mean = 22.325, sample standard deviation = 5.8239, and N = 40 for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.01 significance level. a) Identify the correct alternative hypothesis: μ>21.21μ>21.21 p<21.21p<21.21 p=21.21p=21.21 μ<21.21μ<21.21 p>21.21p>21.21 μ=21.21μ=21.21 Give all answers correct to 3 decimal places. b) The test statistic value is:      c) Using the Traditional method, the critical...
Data: A randomly selected group of 35 adult American men (subscript M) was asked if they...
Data: A randomly selected group of 35 adult American men (subscript M) was asked if they owned a handgun. 20 said yes. A randomly selected group of 40 adult American women (subscript W) was asked if they owned handgun. 21 said yes. Test the claim that the proportion of adult men who own handguns is the same as the proportion of adult women who own handguns. Use α = .05 . Question 7: Express the claim in symbolic form. Group...
The accuracy of a coin-counter machine is gauged to accept nickels with a mean diameter of...
The accuracy of a coin-counter machine is gauged to accept nickels with a mean diameter of millimeters 21.21 mm. A sample of 40 nickles was drawn from a reported defective coin-counter machine located near a school. The sample had a sample mean of 21.209 mm and sample standard deviation 0.007 mm. Test the claim that the mean nickel diameter accepted by this coin-counter machine is greater than 21.21 mm. Test at the 0.01 significance level. (a) Identify the correct alternative...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT