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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

Short Answer

Expert verified

a.There is enough evidence to support the claim that differences between the ages of Best Actresses and Best Actors have a mean less than 0.

b.The confidence interval is equal to (-16.5 years, -2.9 years). The feature that the confidence interval does not contain the value 0 and contains all negative values indicates that the mean value of the differences will always be less than 0.

Step by step solution

01

Given information

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

02

Hypothesis

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.

03

Differences of matched pairs

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

Actress (years)

28

28

31

29

35

26

26

41

30

34

Actor (years)

62

37

36

38

29

34

51

39

37

42

Differences

-34

-9

-5

-9

6

-8

-25

2

-7

-8

\(\sum d = - 97\)

04

Find mean value of the differences of matched pairs

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.

05

Compute the standard deviation

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.

06

Compute the test statistic

The test statistic is equal to:

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

07

Compute degrees of freedom

The degrees of freedom is given by,

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

08

Compute the critical value

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.

09

Decision rule

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.

10

Step 10:Find margin of error to compute the confidence interval

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:

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

11

Confidence interval

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.

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