Bundle: Statistics for Business & Economics, Loose-Leaf Version, 13th + MindTap Business Statistics with XLSTAT, 1 term (6 months) Printed Access Card
Bundle: Statistics for Business & Economics, Loose-Leaf Version, 13th + MindTap Business Statistics with XLSTAT, 1 term (6 months) Printed Access Card
13th Edition
ISBN: 9781337148092
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
Publisher: Cengage Learning
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Chapter 12.2, Problem 16E

The race for the 2013 Academy Award for Actress in a Leading Role was extremely tight, featuring several worthy performances (ABC News online, February 22, 2013). 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.

18–30 31–4 45–58 Over 58
Jessica Chastain 51 50 41 42
Jennifer Lawrence 63 55 37 50
Emmanuelle Riva 15 44 56 74
Quvenzhané Wallis 48 25 22 31
Naomi Watts 36 65 62 33
  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? 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?

a.

Expert Solution
Check Mark
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 (fij) as the row and column total is tabulated below:

ActressAge 18-30Age 31-44Age 45-58Age over 58Total
Jessica Chastain51504142184
Jennifer Lawrence63553750205
Emmanuelle Riva15445674189
Quvenzhane Wallis48252231126
Naomi Watts36656233196
Total213239218230900

Thus, the sample size is 900.

b.

Expert Solution
Check Mark
To determine

Explain whether the respondents favor Ms. Lawrence.

Explanation of Solution

Calculation:

Jennifer Lawrence received the 2013 Academy Award for actress in a leading role for “Silver Lining Playbook”.

The sample proportion of movie fan for Jessica Chastain is,

p1¯=184900=0.2044

The sample proportion of movie fan for Jennifer Lawrence is,

p2¯=205900=0.2278

The sample proportion of movie fan for Emmanuelle Riva is,

p3¯=189900=0.2100

The sample proportion of movie fan for Quvenzhane Wallis is,

p4¯=126900=0.1400

The sample proportion of movie fan for Naomi Watts is,

p5¯=196900=0.2178

It is clear that, the sample proportion for Jennifer Lawrence is highest. Thus, the respondents favor Ms. Lawrence. However, Jessica Chastain, Emmanuelle Riva and Naomi Watts were also favored by almost as many of fans.

c.

Expert Solution
Check Mark
To determine

Perform 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 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 for Actress in a Leading Role is not independent of respondent age.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

 H0:The two cataegorical variables are independent

That is, 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.

Alternative hypothesis:

 Ha:The two cataegorical variables are not independent

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

The row and column total is tabulated below:

ActressAge 18-30Age 31-44Age 45-58Age over 58Total
Jessica Chastain51504142184
Jennifer Lawrence63553750205
Emmanuelle Riva15445674189
Quvenzhane Wallis48252231126
Naomi Watts36656233196
Totals213239218230900

The formula for expected frequency is given below:

eij=(Row i Total)(Coloumn j Total)Total Sample Size

The expected frequency for each category is calculated as follows:

ActressAge 18-30Age 31-44Age 45-58Age over 58
Jessica Chastain(184)(213)900=43.5(184)(239)900=48.9(184)(218)900=44.6(184)(230)900=47
Jennifer Lawrence(205)(213)900=48.5(205)(239)900=54.4(205)(218)900=49.7(205)(230)900=52.4
Emmanuelle Riva(189)(213)900=29.8(189)(239)900=50.2(189)(218)900=45.8(189)(230)900=48.3
Quvenzhane Wallis(126)(213)900=29.8(126)(239)900=33.5(126)(218)900=30.5(126)(230)900=32.2
Naomi Watts(196)(213)900=46.4(196)(239)900=52(196)(218)900=47.5(196)(230)900=50.1

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

χ2=i=1rj=1c(fijeij)2eij where r represent the rth row and c represents the cth column

The value of chi-square test statistic is,

χ2={(5143)243+(5048)248+(4144)244+(4247)247+(6348)248+(5554)254+(3749)249+(5052)252+(1529)229+(4450)250+(5645)245+(7448)248+(4829)229+(2533)233+(2230)230+(3132)232+(3646)246+(6552)252+(6247)247+(3350)250}=(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)=77.74

Thus, the chi-square test statistic is 77.74.

Degrees of freedom:

The degrees of freedom are df=(r1)(c1) for the contingency table of r rows and c columns.

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

Therefore,

df=(51)(41)=12

Level of significance:

The given level of significance is α=0.05.

p-value:

Software procedure:

Step -by-step software procedure to obtain p-value using MINITAB software is as follows:

  • Select Graph > Probability distribution plot > view probability
  • Select chi-square under distribution and enter 12 in degrees of freedom.
  • Choose X-Value and Right Tail for the region of the curve to shade.
  • Enter the X-value as 77.74 under shaded area.
  • Select OK.
  • Output using MINITAB software is given below:

Bundle: Statistics for Business & Economics, Loose-Leaf Version, 13th + MindTap Business Statistics with XLSTAT, 1 term (6 months) Printed Access Card, Chapter 12.2, Problem 16E

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

Rejection rule:

If the p-valueα, then reject the null hypothesis.

Conclusion:

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

That is, p-value(=0)<α(=0.05)

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 for Actress in a Leading Role is not independent of respondent age.

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Chapter 12 Solutions

Bundle: Statistics for Business & Economics, Loose-Leaf Version, 13th + MindTap Business Statistics with XLSTAT, 1 term (6 months) Printed Access Card

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