Question

7) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 .

Express the claim in symbolic form.

Group of answer choices

σ ≠ 6.5

σ ≥ 6.5

σ = 6.5

σ ≤ 6.5

σ > 6.5

σ < 6.5

28) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 &alpha; = .05 .

What is the alternative hypothesis, H1?

Group of answer choices

σ ≥ 6.5

σ < 6.5

σ = 6.5

σ ≤ 6.5

σ ≠ 6.5

σ > 6.5

29) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 .

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

30) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 .

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

31) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 .

What is the statistical conclusion?

Group of answer choices

1. Fail to reject H0
2. Reject H0

32) A retired statistics professor has recorded final exam results for decades. The mean final exam 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 final exam 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 .

State the conclusion in words.

Group of answer choices

a) There is not sufficient evidence to warrant rejection of 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 sample evidence to support 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 .

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

Express the claim in symbolic form.

Group of answer choices

μ < 82.4

μ > 82.4

μ = 82.4

μ ≠ 82.4

μ ≤ 82.4

μ ≥ 82.4

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

What is the alternative hypothesis, H1?

Group of answer choices

μ ≤ 82.4

μ ≠ 82.4

μ > 82.4

μ ≥ 82.4

μ < 82.4

μ = 82.4

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

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

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

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

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

What is the statistical conclusion?

Group of answer choices

1. Reject H0
2. Fail to reject H0

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

State the conclusion in words.

Group of answer choices

1. There is not sufficient sample evidence to support the claim that the current mean is different from 82.4 .
2. There is not sufficient evidence to warrant rejection of the claim that the current mean is different from 82.4 .
3. The sample data support the claim that the current mean is different from 82.4 .
4. There is sufficient evidence to warrant rejection of the claim that the current mean is different from 82.4 .

Answer 1

27)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: σ = 6.5

28)
Alternative Hypothesis, Ha: σ > 6.5

29)

Critical value of Χ^2 are 12.40 and 39.36

30)

Test statistic,
Χ^2 = (n-1)*s^2/σ^2
Χ^2 = (25 - 1)*4.2^2/6.5^2
Χ^2 = 10.0204

31)

reject the null hypothesis.

32)

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

33)

Null Hypothesis: μ = 82.4

34)
Alternative Hypothesis: μ ≠ 82.4

35)

Critical value of z are -1.96 and 1.96.

36)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (80 - 82.4)/(6.5/sqrt(25))
z = -1.846

37)
fail to reject null hypothesis.

38)
There is not sufficient evidence to warrant rejection of the claim that the current mean is different from 82.4 .