Chapter 12.2, Problem 16E

Academy Awards and Movie Fan Sentiment. The race for the 2013 Academy Award for Actress in a Leading Role was extremely tight, featuring several worthy performances. The nominees were Jessica Chastain for Zero Dark Thirty, Jennifer Lawrence for Silver Linings Playbook, Emmanuelle Riva for Amour, Quvenzhané Wallis for Beasts of the Southern Wild, and Naomi Watts for The Impossible. In a survey, movie fans who had seen each of the movies for which these five actresses had been nominated were asked to select the actress who was most deserving of the 2013 Academy Award for Actress in a Leading Role. The responses follow.

1. a. How large was the sample in this survey?
2. b. Jennifer Lawrence received the 2013 Academy Award for Actress in a Leading Role for her performance in Silver Linings Playbook. Did the respondents favor Ms. Lawrence?
3. c. At α = .05, conduct a hypothesis test to determine whether people’s attitude toward the actress who was most deserving of the 2013 Academy Award for Actress in a Leading Role is independent of respondent age. What is your conclusion?

To determine

Find the sample size for the given survey.

Answer to Problem 16E

The sample size is 900.

Explanation of Solution

Calculation:

The given observed frequency $\left(f_{ij}\right)$ as the row and column total is tabulated below:

Actress	Age 18-30	Age 31-44	Age 45-58	Age over 58	Total
JC	51	50	41	42	184
JL	63	55	37	50	205
ER	15	44	56	74	189
QW	48	25	22	31	126
NW	36	65	62	33	196
Total	213	239	218	230	900

Thus, the sample size is 900.

To determine

Explain whether the respondents favor Ms. L.

Explanation of Solution

Calculation:

Actress JL received the 2013 Academy Award for actress in a leading role for movie “SLP”.

The sample proportion of movie fan for Actress JC is,

$\begin{array}{c}\overline{p_1}=\frac{184}{900}\\ =0.2044\end{array}$

The sample proportion of movie fan for Actress JL is,

$\begin{array}{c}\overline{p_2}=\frac{205}{900}\\ =0.2278\end{array}$

The sample proportion of movie fan for Actress ER is,

$\begin{array}{c}\overline{p_3}=\frac{189}{900}\\ =0.2100\end{array}$

The sample proportion of movie fan for Actress QW is,

$\begin{array}{c}\overline{p_4}=\frac{126}{900}\\ =0.1400\end{array}$

The sample proportion of movie fan for Actress NW is,

$\begin{array}{c}\overline{p_5}=\frac{196}{900}\\ =0.2178\end{array}$

It is clear that, the sample proportion for Actress JL is highest. Thus, the respondents favor Ms. JL. However, Actress JC, Actress ER and Actress NW were also favored by almost same number of fans.

To determine

Perform a hypothesis test to determine whether people’s attitude toward the actress who was most deserving of the 2013 Academy Award in a Leading Role, is independent of respondent age at 5% level of significance and draw conclusion of the study.

Answer to Problem 16E

The data provide sufficient evidence to conclude that people’s attitude toward the actress who was most deserving of the 2013 Academy Award in a Leading Role is not independent of respondent age.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

$H_0:\text{The two cataegorical variables are independent}$

That is, people’s attitude toward the actress who was most deserving of the 2013 Academy Award in a Leading Role is independent of respondent age.

Alternative hypothesis:

$H_a:\text{The two cataegorical variables are not independent}$

That is, people’s attitude toward the actress who was most deserving of the 2013 Academy Award in a Leading Role is not independent of respondent age.

The row and column totals are tabulated below:

Actress	Age 18-30	Age 31-44	Age 45-58	Age over 58	Total
Actress JC	51	50	41	42	184
Actress JL	63	55	37	50	205
Actress ER	15	44	56	74	189
Actress QW	48	25	22	31	126
Actress NW	36	65	62	33	196
Totals	213	239	218	230	900

The formula for expected frequency is given below:

$e_{ij}=\frac{\left(\text{Row }i\text{ Total}\right)\left(\text{Coloumn }j\text{ Total}\right)}{\text{Total Sample Size}}$

The expected frequency for each category is calculated as follows:

Actress	Age 18-30	Age 31-44	Age 45-58	Age over 58
Actress JC	$\begin{array}{l}\frac{\left(184\right)\left(213\right)}{900}\\ =43.5\end{array}$	$\begin{array}{l}\frac{\left(184\right)\left(239\right)}{900}\\ =48.9\end{array}$	$\begin{array}{l}\frac{\left(184\right)\left(218\right)}{900}\\ =44.6\end{array}$	$\begin{array}{l}\frac{\left(184\right)\left(230\right)}{900}\\ =47\end{array}$
Actress JL	$\begin{array}{l}\frac{\left(205\right)\left(213\right)}{900}\\ =48.5\end{array}$	$\begin{array}{l}\frac{\left(205\right)\left(239\right)}{900}\\ =54.4\end{array}$	$\begin{array}{l}\frac{\left(205\right)\left(218\right)}{900}\\ =49.7\end{array}$	$\begin{array}{l}\frac{\left(205\right)\left(230\right)}{900}\\ =52.4\end{array}$
Actress ER	$\begin{array}{l}\frac{\left(189\right)\left(213\right)}{900}\\ =44.7\end{array}$	$\begin{array}{l}\frac{\left(189\right)\left(239\right)}{900}\\ =50.2\end{array}$	$\begin{array}{l}\frac{\left(189\right)\left(218\right)}{900}\\ =45.8\end{array}$	$\begin{array}{l}\frac{\left(189\right)\left(230\right)}{900}\\ =48.3\end{array}$
Actress QW	$\begin{array}{l}\frac{\left(126\right)\left(213\right)}{900}\\ =29.8\end{array}$	$\begin{array}{l}\frac{\left(126\right)\left(239\right)}{900}\\ =33.5\end{array}$	$\begin{array}{l}\frac{\left(126\right)\left(218\right)}{900}\\ =30.5\end{array}$	$\begin{array}{l}\frac{\left(126\right)\left(230\right)}{900}\\ =32.2\end{array}$
Actress NW	$\begin{array}{l}\frac{\left(196\right)\left(213\right)}{900}\\ =46.4\end{array}$	$\begin{array}{l}\frac{\left(196\right)\left(239\right)}{900}\\ =52\end{array}$	$\begin{array}{l}\frac{\left(196\right)\left(218\right)}{900}\\ =47.5\end{array}$	$\begin{array}{l}\frac{\left(196\right)\left(230\right)}{900}\\ =50.1\end{array}$

The formula for chi-square test statistic is given as,

${\chi }^2={\displaystyle \sum _{i=1}^r{\displaystyle \sum _{j=1}^c\frac{{\left(f_{ij}-e_{ij}\right)}^2}{e_{ij}}}}$ where $\left(i,j\right)$^th element denotes the frequency corresponding to $\left(i,j\right)$^th cell and r represent the r^th row and c represents the c^th column.

The value of chi-square test statistic is,

$\begin{array}{c}{\chi }^2=\left\{\begin{array}{l}\frac{{\left(51-43\right)}^2}{43}+\frac{{\left(50-48\right)}^2}{48}+\frac{{\left(41-44\right)}^2}{44}+\frac{{\left(42-47\right)}^2}{47}+\frac{{\left(63-48\right)}^2}{48}+\\ \frac{{\left(55-54\right)}^2}{54}+\frac{{\left(37-49\right)}^2}{49}+\frac{{\left(50-52\right)}^2}{52}+\frac{{\left(15-44.7\right)}^2}{44.7}+\frac{{\left(44-50\right)}^2}{50}+\\ \frac{{\left(56-45\right)}^2}{45}+\frac{{\left(74-48\right)}^2}{48}+\frac{{\left(48-29\right)}^2}{29}+\frac{{\left(25-33\right)}^2}{33}+\frac{{\left(22-30\right)}^2}{30}+\\ \frac{{\left(31-32\right)}^2}{32}+\frac{{\left(36-46\right)}^2}{46}+\frac{{\left(65-52\right)}^2}{52}+\frac{{\left(62-47\right)}^2}{47}+\frac{{\left(33-50\right)}^2}{50}\end{array}\right\}\\ =\left(\begin{array}{c}1.28+0.03+0.29+0.54+4.32+0.01+3.23+0.11+19.76+0.76+\\ 2.28+13.67+11.08+2.14+2.38+0.04+2.33+3.22+4.44+5.83\end{array}\right)\\ =77.74\end{array}$

Thus, the chi-square test statistic is 77.74.

Degrees of freedom:

The degrees of freedom are $df=\left(r-1\right)\left(c-1\right)$ for the contingency table of r rows and c columns.

In the given problem $r=5$ and $c=4$.

Therefore,

$\begin{array}{c}df=\left(5-1\right)\left(4-1\right)\\ =12\end{array}$

Level of significance:

The given level of significance is $\alpha =0.05$.

p-value:

Software procedure:

Step -by-step software procedure to obtain p-value using EXCEL software:

• Open an EXCEL sheet and select cell A1.
• In cell A1 enter the formula =CHISQ.DIST.RT(77.74,12).
• Press Enter.
• Output using EXCEL software is given below:

From the EXCEL output, the p-value is 0.

Rejection rule:

• If the $p\text{-value}\le \alpha$, then reject the null hypothesis.
• Otherwise, failed to reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$

Thus, the decision is “reject the null hypothesis”.

Therefore, the data provide sufficient evidence to conclude that column variable is not independent of row variable. That is, there is an association between column and row variable.

Thus, the data provide sufficient evidence to conclude that people’s attitude toward the actress who was most deserving of the 2013 Academy Award in a Leading Role is not independent of respondent age.