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Part 1
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 nequals82​, xequals35.3 ​years, and sequals11.1 years. Use a 0.01 significance level to test the claim that the mean age of actresses when they win an acting award is 33 years.
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Part 1
What are the​ hypotheses?
A.
Upper H 0​: muequals33 years
Upper H 1​: mugreater than or equals33 years
B.
Upper H 0​: muequals33 years
Upper H 1​: muless than33 years
C.
Upper H 0​: munot equals33 years
Upper H 1​: muequals33 years
D.
Upper H 0​: muequals33 years
Upper H 1​: munot equals33 years
Your answer is correct.
Part 2
Identify the test statistic.


enter your response here ​(Round to three decimal places as​ needed.)

### Part 1: Hypotheses

The hypotheses for this test can be stated as follows:

- **Null Hypothesis (\(H_0\))**: \(\mu = 33\) years
- **Alternative Hypothesis (\(H_1\))**: \(\mu \neq 33\) years

These hypotheses are consistent with the claim that we are testing whether the mean age of actresses when they win an acting award is different from 33 years. Therefore, the correct answer is:

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

### Part 2: Identify the Test Statistic

To find the test statistic for a hypothesis test concerning a population mean, we use the formula for the t-test statistic:

\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]

Given the following values:
- \(\bar{x} = 35.3\) years (sample mean)
- \(\mu = 33\) years (population mean under the null hypothesis)
- \(s = 11.1\) years (sample standard deviation)
- \(n = 82\) (sample size)

We can substitute these values into the formula:

\[
t = \frac{35.3 - 33}{11.1 / \sqrt{82}}
\]

First, calculate the standard error:

\[
SE = \frac{11.1}{\sqrt{82}} \approx \frac{11.1}{9.055} \approx 1.226
\]

Next, calculate the t-test statistic:

\[
t = \frac{35.3 - 33}{1.226} \approx \frac{2.3}{1.226} \approx 1.876
\]

So, the test statistic (rounded to three decimal places) is:

\[
t \approx 1.876
\]

### Conclusion

With the test statistic calculated, the next steps would involve comparing this t-value to the critical value from the t-distribution table at the \(0.01\) significance level for a two-tailed test, or using a p-value approach to determine if we can reject the null hypothesis. 

Would you like to proceed with calculating the p-value and making the final conclusion?

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Here are 8 additional questions that expand on this information:

1. How do you determine the degrees of freedom for the t-test in this scenario?
2. What are the critical t-values for a two-tailed test at the 0.01 significance level with 81 degrees of freedom?
3. How do you calculate the p-value from the t-test statistic?
4. What decision rule is used to reject or fail to reject the null hypothesis based on the p-value?
5. Can you explain what a Type I error is in the context of this hypothesis test?
6. How would the test conclusions change if the significance level was 0.05 instead of 0.01?
7. What assumptions must be met for the t-test to be valid in this context?
8. How would you interpret the results of this test in the context of the original claim about actresses' ages?

### Tip:
When conducting hypothesis tests, always ensure your sample size is sufficient to provide reliable results and check if the assumptions of the test are satisfied.

Explore hypothesis testing using the P-value method to test the claim about the mean age of actresses when they win an acting award. Learn about the null and alternative hypotheses, calculate the test statistic, and interpret the results. Includes a step-by-step example with n = 82, x̄ = 35.3 years, s = 11.1 years, μ = 33 years, significance level = 0.01.

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