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 left parenthesis years right parenthesis 27 29 32 27 37 28 26 45 28 36 Actor left parenthesis years right parenthesis 62 37 38 37 30 36 48 41 39 42 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, mu Subscript 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?

Answer 1

The null hypothesis is $\fbox{\begin{minispace}\\H_{0}\text{( null hypothesis)}: \mu_{d} \geq0\end{minispace}}$ and the alternative hypothesis $\fbox{\begin{minispace}\\H_{1}\text{( alternative hypothesis)}: \mu_{d}$ and the mean difference is greater than zero and we can conclude that the Best Actresses are generally younger than Best Actors.

Further explanation:

Part (a)

$\fbox{\begin{minispace}\\H_{0}\text{( null hypothesis)}: \mu_{d} \geq0\end{minispace}}$

$\fbox{\begin{minispace}\\H_{1}\text{( alternative hypothesis)}: \mu_{d}$

Part (b)

To find the difference between the Actors age and Actresses age. solve as below.

Difference = Actors age - Actresses age

Kindly refer to the Table for the difference between the Actors and Actress's age.

The number of observations are $10$.

Level of significance = $5\%$

The sum of difference = $95$

So mean of the difference can be calculated as,

The mean of difference = $\dfrac{95}{10}$

The mean of difference $\bar{X}$ = $9.5$

The standard deviation s is,

$s =11.98$

$\fbox{\begin{minispace}\\H_{0}\text{( null hypothesis)}: \mu_{d} \geq0\end{minispace}}$

$\fbox{\begin{minispace}\\H_{1}\text{( alternative hypothesis)}: \mu_{d}$

Under $H_{0}$ Test Statistic Value (TSV) can be calculated as,

TSV = $\dfrac{\bar{X}-\mu_{d}}{\dfrac{s}{\sqrt{n}}}$

TSV = $\dfrac{{9.5}-{0}}{\dfrac{11.98}{\sqrt{10}}}$

TSV = $\dfrac{{9.5}-{0}}{\dfrac{11.98}{\sqrt{10}}}$

TSV = $\dfrac{9.5}{3.79}$

TSV = $2.51$

Critical value can be obtained from the table as,

CV = $t_{10-1,5\%}$

CV = $t_{9,5\%}$

CV = $1.833$

Critical value is less than the test statistic value. Therefore, reject the null hypothesis at $5\%$ level of significance.

Therefore, the mean difference is greater than zero and we can conclude that the Best Actresses are generally younger than Best Actors.

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Answer Details:

Grade: College Statistics

Subject: Statistics

Chapter: Hypothesis Testing

Keywords:

Probability, Statistics, Best Actors, Best Actresses, Mean difference, Hypothesis testing, Null hypothesis, Alternative hypothesis, t-test, Level of significance, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance, Critical Value, Test Statistic Value.