Oscar Hypothesis Test

a.Example 1 on page 444 in this section used only five pairs of data from Data Set 14 “OscarWinner Age” in Appendix B. Repeat the hypothesis test of Example 1 using the data given below. Use a 0.05 significance level as in Example 1.

b.Construct the confidence interval that could be used for the hypothesis test described inpart (a). What feature of the confidence interval leads to the same conclusion reached in part (a)?

Actress (years)	28	28	31	29	35	26	26	41	30	34
Actor (years)	62	37	36	38	29	34	51	39	37	42

The ages of 10 pairs of actors and actresses are recorded when they won the Oscar. Therefore, the sample size\(n = 10\). The level of significance \(\alpha = 0.05\). It is claimed that the differences in the population of all pairs of ages is less than 0

The following hypotheses are noted:

Null Hypothesis: The differences in the population of all pairs of ages is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The differences in the population of all pairs of ages is less than 0.

\({H_1}:{\mu _d} < 0\)

Since the alternative hypothesis contains less than a symbol, it is a left-tailed test.

The following table shows the differences in the ages of actors and actresses for each pair:

Themean value of the differences between the age of actors and actresses:

\(\begin{array}{c}\bar d = \frac{{\sum d }}{n}\\ = \frac{{ - 97}}{{10}}\\ = - 9.7\end{array}\)

Thus, the mean value is -0.28 years.

Thestandard deviation of the differences between the age of actors and actresses is equal to:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 34 - \left( { - 9.7} \right)} \right)}^2} + {{\left( { - 9 - \left( { - 9.7} \right)} \right)}^2} + ....... + {{\left( { - 8 - \left( { - 9.7} \right)} \right)}^2}}}{{10 - 1}}} \\ = 11.76\end{array}\)

Thus, the standard deviation is equal to 11.76 years.

The test statistic is equal to:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 9.7 - 0}}{{\frac{{11.76}}{{\sqrt {10} }}}}\\ = - 2.609\end{array}\)

The degrees of freedom is given by,

\(\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\)

Referring to the t-distribution table, use the column for 0.05 (Area in One Tail),and use the row with 9 degrees of freedom to find the critical value.

\(\begin{array}{c}{t_{crit}} = {t_\alpha },df\\ = {t_{0.05}},9\\ = - 1.833\end{array}\)

Thus, the critical value is -1.833.

a.

For the left tailed test, if the calculated value is less than the critical value, reject the null hypothesis. In this scenario,\( - 2.57 < - 1.83\). Hence, reject the null hypothesis.

There is sufficient evidence to support the claim that the age differences between the actresses and actors when they won the Oscar have a mean of less than 0.

If the level of significance is 0.05 for one-tailed test, then use 90% confidence level to construct the confidence interval.

The margin of error is equal to:

\(\)\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = 1.833 \times \frac{{11.76}}{{\sqrt {10} }}\\ = 6.815\end{array}\)

b.

The formula for the confidence interval is equal to:

\(\begin{array}{c}CI = \bar d - E < {\mu _d} < \bar d + E\\ = - 9.7 - 6.815 < {\mu _d} < - 9.7 + 6.815\\ = - 16.5 < {\mu _d} < - 2.9\end{array}\)

Thus, the confidence interval is equal to (-16.5 years,-2.9 years).

Here clearly, one can see that 0 is not included in the interval. So, it is not possible that the mean of the differences will be equal to zero.

Hencethere is sufficient evidence to support the claim that age differences between the best actresses and best actors when they won the Oscar have a mean less than 0.