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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.

\begin{tabular}{lllllllllll}
\hline Actress (years) & 30 & 30 & 30 & 29 & 35 & 27 & 26 & 40 & 30 & 33 \\
\hline Actor (years) & 64 & 40 & 36 & 36 & 29 & 35 & 52 & 37 & 34 & 38 \\
\hline
\end{tabular}

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

(Type integers or decimals. Do not round.)

Sagot :

GinnyAnswer
### Step-by-Step Solution

#### (a) Hypothesis Testing

We are given the ages of actresses and actors when they won awards and need to test if there is a significant difference in their ages, particularly if actresses are generally younger than actors. Here we'll conduct a hypothesis test with a significance level of 0.05.

1. State the Hypotheses:

- Null Hypothesis ([tex]\( H_0 \)[/tex]): The mean difference in ages ([tex]\( \mu_d \)[/tex]) between actresses and actors is greater than or equal to 0. This implies actresses are not significantly younger than actors.
[tex]\[ H_0: \mu_d \geq 0 \][/tex]

- Alternative Hypothesis ([tex]\( H_1 \)[/tex]): The mean difference in ages ([tex]\( \mu_d \)[/tex]) is less than 0. This implies actresses are significantly younger than actors.
[tex]\[ H_1: \mu_d < 0 \][/tex]

- Here, [tex]\( \mu_d \)[/tex] represents the mean difference where each difference [tex]\( d \)[/tex] is calculated by [tex]\( \text{actress's age} - \text{actor's age} \)[/tex].

2. Sample Data:

The ages of actresses and actors are given as follows:

- Ages of Actresses: [tex]\( [30, 30, 30, 29, 35, 27, 26, 40, 30, 33] \)[/tex]
- Ages of Actors: [tex]\( [64, 40, 36, 36, 29, 35, 52, 37, 34, 38] \)[/tex]

3. Calculate the Differences:

Differences [tex]\( d \)[/tex] are calculated for matched pairs:
[tex]\[ [30-64, 30-40, 30-36, 29-36, 35-29, 27-35, 26-52, 40-37, 30-34, 33-38] \][/tex]

4. Sample Mean of Differences and Test Statistics:

From the result provided, the numerical values are:

- Mean of Differences ([tex]\( \bar{d} \)[/tex]): [tex]\( -9.1 \)[/tex]
- Test Statistic ([tex]\( t \)[/tex]-statistic): [tex]\( -2.359 \)[/tex]
- P-value: [tex]\( 0.0213 \)[/tex]

5. Comparison with Critical Value:

Since we are conducting a one-sample t-test with a one-tailed hypothesis (left-tailed), we compare the p-value to our significance level ([tex]\( \alpha = 0.05 \)[/tex]).

6. Make Decision:

- If the p-value < 0.05, we reject the null hypothesis.
- Here, [tex]\( 0.0213 < 0.05 \)[/tex], so we reject the null hypothesis.

7. Conclusion:

At a 0.05 significance level, there is sufficient evidence to support the claim that Best Actresses are generally younger than Best Actors.

### Answers for the Hypotheses

[tex]\[ \begin{array}{l} H_0: \mu_d \geq 0 \\ H_1: \mu_d < 0 \end{array} \][/tex]
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