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The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, along with the ages of actors when they won in the category of Best Actor. The ages are matched according to the year that the awards were presented. Complete parts (a) and (b) below.

[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\text{Actress (years)} & 29 & 31 & 30 & 29 & 34 & 28 & 28 & 44 & 28 & 32 \\
\hline
\text{Actor (years)} & 68 & 40 & 37 & 42 & 28 & 32 & 50 & 37 & 35 & 44 \\
\hline
\end{array}
\][/tex]

a. Use the sample data with a 0.05 significance level to test the claim that for the population of ages of Best Actresses and Best Actors, the differences have a mean less than 0 (indicating that the Best Actresses are generally younger than Best Actors).

In this example, [tex]\mu_d[/tex] is the mean value of the differences [tex]d[/tex] for the population of all pairs of data, where each individual difference [tex]d[/tex] is defined as the actress's age minus the actor's age. What are the null and alternative hypotheses for the hypothesis test?

[tex]\[
\begin{array}{l}
H_0: \mu_{d} = 0 \text{ year(s)} \\
H_1: \mu_{d} \ \textless \ 0 \text{ year(s)}
\end{array}
\][/tex]

Identify the test statistic.

[tex]t = \square[/tex] (Round to two decimal places as needed.)

Answer :

To solve the problem of testing the claim that Best Actresses are generally younger than Best Actors when they receive their awards, we'll follow these steps:

### Step 1: State the Hypotheses

- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean of the differences ([tex]\(\mu_d\)[/tex]) between ages of actresses and actors is equal to 0. This suggests there is no age difference, on average.
[tex]\[
H_0: \mu_d = 0
\][/tex]

- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean of the differences ([tex]\(\mu_d\)[/tex]) is less than 0, suggesting actresses are generally younger than actors.
[tex]\[
H_1: \mu_d < 0
\][/tex]

### Step 2: Calculate the Differences

For each pair, find the age difference between the actress and the actor:
- Calculate the differences: [29 - 68, 31 - 40, 30 - 37, 29 - 42, 34 - 28, 28 - 32, 28 - 50, 44 - 37, 28 - 35, 32 - 44]

### Step 3: Compute the Mean and Standard Deviation of Differences

- Mean Difference ([tex]\(\bar{d}\)[/tex]): The average of the differences.
[tex]\[
\bar{d} = -10.0
\][/tex]

- Standard Deviation ([tex]\(s_d\)[/tex]): This measures the variability of the differences.
[tex]\[
s_d = 13.32
\][/tex]

### Step 4: Calculate the Test Statistic

The test statistic for a paired t-test is calculated using:
[tex]\[
t = \frac{\bar{d}}{s_d / \sqrt{n}}
\][/tex]

Where:
- [tex]\(\bar{d} = -10.0\)[/tex] is the mean difference.
- [tex]\(s_d = 13.32\)[/tex] is the standard deviation.
- [tex]\(n = 10\)[/tex] is the number of pairs.

Substituting the values:
[tex]\[
t = \frac{-10.0}{13.32 / \sqrt{10}} \approx -2.37
\][/tex]

### Conclusion

The test statistic [tex]\(t \approx -2.37\)[/tex] helps us determine if we reject the null hypothesis based on the critical value for a one-tailed test with a 0.05 significance level. If the calculated [tex]\(t\)[/tex]-value falls in the critical region (less than the critical value), we reject the null hypothesis and support the claim that Best Actresses are generally younger than Best Actors.

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