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 

nequals=8282​, 

xequals=35.335.3 

​years, and 

sequals=11.111.1 

years. Use a 

0.010.01 

significance level to test the claim that the mean age of actresses when they win an acting award is 

3333 

years.

Question content area bottom

Part 1

What are the​ hypotheses?

A.

Upper H 0H0​: 

muμequals=3333 

years

Upper H 1H1​: 

muμgreater than or equals≥3333 

years

B.

Upper H 0H0​: 

muμequals=3333 

years

Upper H 1H1​: 

muμless than<3333 

years

C.

Upper H 0H0​: 

muμnot equals≠3333 

years

Upper H 1H1​: 

muμequals=3333 

years

D.

Upper H 0H0​: 

muμequals=3333 

years

Upper H 1H1​: 

muμnot equals≠3333 

years

Your answer is correct.

Part 2

Identify the test statistic.

tequals=1.8761.876 

​(Round to three decimal places as​ needed.)

Part 3

Identify the​ P-value.

The​ P-value is 

enter your response here. 

​(Round to four decimal places as​ needed.) 

### Part 1: Hypotheses

The hypotheses for this test are:
- **Null Hypothesis (\(H_0\))**: \(\mu = 33\) years
- **Alternative Hypothesis (\(H_1\))**: \(\mu \neq 33\) years

The correct answer is:

**D. \( H_0: \mu = 33 \) years, \( H_1: \mu \neq 33 \) years**

### Part 2: Test Statistic

The test statistic has already been calculated as \( t = 1.876 \).

### Part 3: Identify the P-value

To find the p-value, we need to use the t-distribution with \( n - 1 = 82 - 1 = 81 \) degrees of freedom. Since this is a two-tailed test, we need to find the probability that the absolute value of the t-statistic is greater than or equal to 1.876.

Using a t-distribution table or a calculator for the t-distribution:

\[
P(\text{abs}(t) \geq 1.876) = 2 \times P(t \geq 1.876 \text{ with } 81 \text{ degrees of freedom})
\]

Using a t-distribution calculator, we find that:

\[
P(t \geq 1.876 \text{ with } 81 \text{ degrees of freedom}) \approx 0.0333
\]

Therefore, the p-value is:

\[
\text{P-value} = 2 \times 0.0333 \approx 0.0666
\]

### Conclusion

Since the p-value (\(0.0666\)) is greater than the significance level (\(\alpha = 0.01\)), we fail to reject the null hypothesis. This means there is not enough evidence to support the claim that the mean age of actresses when they win an acting award is different from 33 years.

### Summary
- **Null Hypothesis (\(H_0\))**: \(\mu = 33\) years
- **Alternative Hypothesis (\(H_1\))**: \(\mu \neq 33\) years
- **Test Statistic**: \( t = 1.876 \)
- **P-value**: \( 0.0666 \)

**Conclusion**: Fail to reject the null hypothesis. There is not enough evidence at the 0.01 significance level to support the claim that the mean age of actresses when they win an acting award is different from 33 years.

Would you like to explore further details or have any questions?

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Here are 8 additional questions that expand on this information:

1. How does the significance level affect the conclusion of a hypothesis test?
2. What are the implications of failing to reject the null hypothesis in this context?
3. How would you calculate the confidence interval for the mean age of actresses when they win an acting award?
4. What are Type I and Type II errors in the context of this hypothesis test?
5. How does the sample size influence the reliability of the hypothesis test?
6. What other statistical tests could be used to analyze this data?
7. How would the test results change if the sample standard deviation were different?
8. Can you explain the importance of the p-value in hypothesis testing?

### Tip:
When performing hypothesis tests, always ensure you understand the assumptions and limitations of the test you are using, and interpret the results in the context of your data and research question.

Explore how to test the claim about the mean age of actresses when they win acting awards using the P-value method. Learn about hypothesis testing, the one-sample t-test, and interpreting P-values with real statistical data.

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