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Assume that a simple random sample has been selected and test the given claim. Use the P-value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. The ages of actresses when they won an acting award is summarized by the statistics n=77, x=35.4 years, and s=11.7 years.

Use a 0.05 significance level to test the claim that the mean age of actresses when they win an acting award is 33 years.
a. What are the hypotheses?
b. Identify the test statistic.
c. Identify the P-value.
d. State the final conclusion that addresses the original claim.
A. Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.
B. Reject H0. There is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.
C. Fail to reject H0. There is sufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.
D. Reject H0. There is sufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.

User Lpacheco
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1 Answer

11 votes

Answer:

a. H0: u ≠33 against Ha : u= 33

b. Z= x`- u / s/ √n, z= 1.85

c. The P-Value is 0.071075.

The result is not significant at p < .05.

d. Option A is correct.

A. Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.

Explanation:

Part a:

The null and alternate hypotheses are

H0: u ≠33 That is the mean age of actresses when they win an acting award is not equal to 33 years.

against the claim

Ha : u= 33 That is the mean age of actresses when they win an acting award is 33 years.

The central limit theorem allows us to assume that the sampling distribution of x` is approximately normal with the mean u and a standard deviation S/ √n.

Part b.

The test statistic would be

Z= x`- u / s/ √n

So we have

n= 77 , x= 35.4, s= 11.7

Z= 35.4- 33/ 11.7/√77

z= 2.4/1.333

z= 1.805

Now z α/2 = ± 1.96 for two tailed test.

Part d.

The rejection region will be

z< - z ∝/2 and z > z ∝/2

1.805 < 1.96

Since the calculated value of z lies in the rejection region we fail to reject H0 and conclude there is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 33 years.

Part c.

The p- value is the probabilities between which the value of test statistic lies.

Using the calculator

The P-Value is 0.071075.

The result is not significant at p < .05

Assume that a simple random sample has been selected and test the given claim. Use-example-1
User TimSPQR
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