Question

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.

actress (years) | actor (years)
--- | ---
27 | 64
26 | 33
33 | 34
31 | 38
37 | 31
25 | 35
26 | 55
39 | 39
33 | 39
35 | 38

a. use the sample data with a 0.01 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, \\( \mu_d \\) is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the actress’s age minus the actor’s age.
what are the null and alternative hypotheses for the hypothesis test?
\\( h_0: \mu_d \\) \\( \square \\) year(s)
\\( h_1: \mu_d \\) \\( \square \\) year(s)
(type integers or decimals. do not round.)
identify the test statistic.
\\( t = \square \\) (round to two decimal places as needed.)
identify the p - value.
p - value = \\( \square \\) (round to three decimal places as needed.)
what is the conclusion based on the hypothesis test?
since the p - value is \\( \square \\) the significance level, \\( \square \\) the null hypothesis. there \\( \square \\) sufficient evidence to support the claim that actresses are generally younger when they won the award than actors.

b. construct the confidence interval that could be used for the hypothesis test described in part (a). what feature of the confidence interval leads to the same conclusion reached in part (a)?
the confidence interval is \\( \square \\) year(s) < \\( \mu_d \\) < \\( \square \\) year(s).
(round to one decimal place as needed.)
what feature of the confidence interval leads to the same conclusion reached in part (a)?
since the confidence interval contains \\( \square \\) the null hypothesis.

Explanation:

Step1: Define Differences

First, calculate the differences \( d = \text{Actress Age} - \text{Actor Age} \) for each pair:
- \( 27 - 64 = -37 \)
- \( 26 - 33 = -7 \)
- \( 33 - 34 = -1 \)
- \( 31 - 38 = -7 \)
- \( 37 - 31 = 6 \)
- \( 25 - 35 = -10 \)
- \( 26 - 55 = -29 \)
- \( 39 - 39 = 0 \)
- \( 33 - 39 = -6 \)
- \( 35 - 38 = -3 \)

Step2: Calculate \( \bar{d} \) and \( s_d \)

Compute the mean of differences \( \bar{d} \):
\[ \bar{d} = \frac{-37 - 7 - 1 - 7 + 6 - 10 - 29 + 0 - 6 - 3}{10} = \frac{-94}{10} = -9.4 \]
Compute the standard deviation of differences \( s_d \). First, find squared deviations from \( \bar{d} \):
- \( (-37 + 9.4)^2 = (-27.6)^2 = 761.76 \)
- \( (-7 + 9.4)^2 = (2.4)^2 = 5.76 \)
- \( (-1 + 9.4)^2 = (8.4)^2 = 70.56 \)
- \( (-7 + 9.4)^2 = (2.4)^2 = 5.76 \)
- \( (6 + 9.4)^2 = (15.4)^2 = 237.16 \)
- \( (-10 + 9.4)^2 = (-0.6)^2 = 0.36 \)
- \( (-29 + 9.4)^2 = (-19.6)^2 = 384.16 \)
- \( (0 + 9.4)^2 = (9.4)^2 = 88.36 \)
- \( (-6 + 9.4)^2 = (3.4)^2 = 11.56 \)
- \( (-3 + 9.4)^2 = (6.4)^2 = 40.96 \)

Sum of squared deviations: \( 761.76 + 5.76 + 70.56 + 5.76 + 237.16 + 0.36 + 384.16 + 88.36 + 11.56 + 40.96 = 1606.4 \)

Sample standard deviation \( s_d \):
\[ s_d = \sqrt{\frac{1606.4}{10 - 1}} = \sqrt{\frac{1606.4}{9}} \approx 13.33 \]

Step3: State Hypotheses

Null hypothesis \( H_0: \mu_d = 0 \), Alternative hypothesis \( H_1: \mu_d < 0 \) (since we test if actresses are younger, \( d = \text{actress} - \text{actor} < 0 \) on average).

Step4: Calculate Test Statistic

Test statistic \( t \) for paired t - test:
\[ t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}} = \frac{-9.4 - 0}{13.33 / \sqrt{10}} \approx \frac{-9.4}{4.21} \approx -2.23 \]

Step5: Find P - value

Degrees of freedom \( df = n - 1 = 9 \). For \( t = -2.23 \) (left - tailed test), the P - value is the probability that \( T < -2.23 \) with \( df = 9 \). Using t - table or calculator, P - value \( \approx 0.026 \) (rounded to three decimal places).

Step6: Conclusion for (a)

Since \( \alpha = 0.01 \), and P - value \( = 0.026 > 0.01 \), we fail to reject \( H_0 \). There is not sufficient evidence at \( \alpha = 0.01 \) to support the claim.

Step7: Confidence Interval for (b)

For a 99% confidence interval (since \( \alpha = 0.01 \), confidence level \( = 1 - \alpha = 0.99 \)) for paired differences:
\[ \text{Confidence Interval} = \bar{d} \pm t_{\alpha/2, df} \cdot \frac{s_d}{\sqrt{n}} \]
\( t_{0.005, 9} = 3.250 \) (from t - table)
\[ \text{Margin of Error} = 3.250 \cdot \frac{13.33}{\sqrt{10}} \approx 3.250 \cdot 4.21 \approx 13.68 \]
\[ \text{Confidence Interval} = -9.4 \pm 13.68 = (-23.08, 4.28) \]
Rounding to one decimal place: \( -23.1 < \mu_d < 4.3 \) (wait, correction: earlier \( \bar{d} \) calculation was wrong. Wait, recalculating \( \bar{d} \):
Wait, let's recalculate differences correctly:
1. \( 27 - 64 = -37 \)
2. \( 26 - 33 = -7 \)
3. \( 33 - 34 = -1 \)
4. \( 31 - 38 = -7 \)
5. \( 37 - 31 = 6 \)
6. \( 25 - 35 = -10 \)
7. \( 26 - 55 = -29 \)
8. \( 39 - 39 = 0 \)
9. \( 33 - 39 = -6 \)
10. \( 35 - 38 = -3 \)

Sum: \( -37 -7 = -44; -44 -1 = -45; -45 -7 = -52; -52 +6 = -46; -46 -10 = -56; -56 -29 = -85; -85 +0 = -85; -85 -6 = -91; -91 -3 = -94 \). \( \bar{d} = -94/10 = -9.4 \) (correct). \( s_d \) calculation: sum of squared deviations:
\( (-37 + 9.4)^2 = (-27.6)^2 = 761.76 \)
\( (-7 + 9.4)^2 = 2.4^2 = 5.76 \)
\( (-1 + 9.4)^2 = 8.4^2 = 70.56 \)
\( (-7 + 9.4)^2 = 2.4^2 = 5.76 \)
\( (6 + 9.4)^2 = 15.4^2 = 237.16 \)
\( (-10 + 9.4)^2 = (-0.6)^2 = 0.36 \)
\( (-29 + 9.4)^2 = (-19.6)^2 = 384.16 \)
\( (0 + 9.4)^2 = 9.4^2 = 88.36 \)
\( (-6 + 9.4)^2 = 3.4^2 = 11.56 \)
\( (-3 + 9.4)^2 = 6.4^2 = 40.96 \)
Sum: \( 761.76+5.76 = 767.52; +70.56 = 838.08; +5.76 = 843.84; +237.16 = 1081; +0.36 = 1081.36; +384.16 = 1465.52; +88.36 = 1553.88; +11.56 = 1565.44; +40.96 = 1606.4 \). \( s_d=\sqrt{1606.4/9}=\sqrt{178.49}\approx13.36 \) (corrected \( s_d \)). Then margin of error: \( 3.250\times(13.36/\sqrt{10})\approx3.250\times4.22\approx13.72 \). Confidence interval: \( -9.4\pm13.72=(-23.12, 4.32) \), rounded to one decimal: \( -23.1 < \mu_d < 4.3 \). The confidence interval contains 0 (since lower bound - 23.1 and upper bound 4.3 includes 0), which means we fail to reject \( H_0 \) (same conclusion as (a) because 0 is in the interval, so no significant difference at \( \alpha = 0.01 \)).

Answer:

Part (a)

• Null Hypothesis \( H_0: \mu_d = 0 \) (years), Alternative Hypothesis \( H_1: \mu_d < 0 \) (years)
• Test Statistic \( t \approx -2.23 \)
• P - value \( \approx 0.026 \)
• Conclusion: Fail to reject \( H_0 \). There is not sufficient evidence at \( \alpha = 0.01 \) to support the claim that actresses are generally younger when they won the award than actors.

Part (b)

• Confidence Interval: \( -23.1 \) years \( < \mu_d < 4.3 \) years
• Feature: The confidence interval contains 0, so we fail to reject the null hypothesis (same as part (a)).