Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).) Bad Stuff in Children’s Movies Data Set 11 “Alcohol and Tobacco in Movies” in Appendix B includes lengths of times (seconds) of tobacco use shown in animated children’s movies. For the Disney movies, n = 33,\(\bar x\)= 61.6 sec, s = 118.8 sec. For the other movies, n = 17,\(\bar x\)= 49.3 sec, s = 69.3 sec. The sorted times for the non-Disney movies are listed below.

a. Use a 0.05 significance level to test the claim that Disney animated children’s movies and other animated children’s movies have the same mean time showing tobacco use.

b. Construct a confidence interval appropriate for the hypothesis test in part (a).

c. Conduct a quick visual inspection of the listed times for the non-Disney movies and comment on the normality requirement. How does the normality of the 17 non-Disney times affect the results?

0 0 0 0 0 0 1 5 6 17 24 55 91 117 155 162 205

Short answer

a.There is not enough evidence to reject the claim that Disney animated children’s movies and other animated children’s movies have the same mean time showing tobacco use.

b.The 95% confidence interval is equal to(-44.2,-68.8).

c.The following normal quantile plot is constructed for the sample of times of tobacco use for the given 17 non-Disney movies:

The given sample of times of tobacco use for the non-Disney movies does not appear to be normally distributed.

The property of the sample not being from a normally distributed population does not majorly affect the results of the test.

Step-by-step solution

Given information

A sample of Disney movies is considered showing the duration of tobacco use in the movies. Another sample of other movies is considered showing the duration of tobacco use in the movies.

Formulation of the hypotheses

Null hypothesis:Disney animated children’s movies and other animated children’s movies have the same mean time showing tobacco use.

\({H_0}\):\({\mu _1} = {\mu _2}\).

Alternative hypothesis: Disney animated children’s movies and other animated children’s movies have different mean time showing tobacco use.

\({H_1}\):\({\mu _1} \ne {\mu _2}\).

Important sample statistics

Let\({n_1}\)denote the sample size corresponding to Disney movies.

\({n_1} = 33\)

Let\({n_2}\)denote the sample size corresponding to other movies.

\({n_1} = 17\)

Let\({\bar x_1}\)denote the mean duration of tobacco use shown in Disney movies.

\({\bar x_1} = 61.6\)

Let\({\bar x_2}\)denote the mean duration of tobacco use shown in other movies.

\({\bar x_2} = 49.3\)

Let\({s_1}\)denote the standard deviation of the duration of tobacco use shown in Disney movies.

\({s_1} = 118.8\)

Let\({s_2}\)denote the standard deviation of the duration of tobacco use shown in other movies.

\({s_2} = 69.3\)

Computation of the test statistic

Under null hypothesis,\({\mu _1} - {\mu _2} = 0\).

The test statistic is equal to:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {61.6 - 49.3} \right) - \left( 0 \right)}}{{\sqrt {\frac{{{{118.8}^2}}}{{33}} + \frac{{{{69.3}^2}}}{{17}}} }}\\ = 0.4616\end{array}\)

Thus, the value of the test statistic is 0.4616.

Computation of critical value

Degrees of freedom: The smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\)is considered as the degrees of freedom.

\(\begin{array}{c}\left( {{n_1} - 1} \right) = \left( {33 - 1} \right)\\ = 32\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = \left( {17 - 1} \right)\\ = 16\end{array}\)

Thus, the degree of freedom is the minimum of (32,16) which is equal to 16.

Now see the t-distribution table for the two-tailed test at a 0.05 level of significance and for 16 degrees of freedom. The critical values are -2.120 and 2.120.

The corresponding p-value for the test statistic value of 0.4616 is equal to 0.6506.

The value of the test statistic lies between the two critical values, and the p-value is greater than 0.05.

Therefore, the null hypothesis fails to be rejected at a 0.05 significance level.

Conclusion

a.

There is not enough evidence to reject the claim that Disney animated children’s movies and other animated children’s movies have the same mean time showing tobacco use.

Formula of the confidence interval

The given formula is used to compute confidence interval:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

Where,

\({\bar x_1}\)is sample mean of the first sample.

\({\bar x_2}\)is sample mean of the second sample.

E is the margin of error

Computation of the margin of error

If the hypothesis test is two-tailed and the level of significance is equal to 0.05, then the confidence level to construct the confidence interval is 95%.

The margin of error is equal to:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = 2.120 \times \sqrt {\frac{{{{\left( {118.8} \right)}^2}}}{{33}} + \frac{{{{\left( {69.3} \right)}^2}}}{{17}}} \\ = 56.50\end{array}\)

The margin of error is 56.50.

Computation of confidence interval

b.

The 95% confidence interval is equal to:

\(\begin{array}{c}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\\left( {61.6 - 49.3} \right) - 56.50 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {61.6 - 49.3} \right) + 56.50\\ - 44.2 < \left( {{\mu _1} - {\mu _2}} \right) < 68.8\end{array}\)

Therefore, the confidence interval is equal to (-44.2, 68.8).

Normal quantile plot

c.

Follow the given steps to construct a normal quantile plot for the given duration of tobacco use for non-Disney movies.

Arrange the given values in ascending order as shown:

0	0	0	0	0	0	1	5	6
17	24	55	91	117	155	162	205

Compute the cumulative areas to the left for each sample value as follows:

Sample Value	Areas to the left
0	\(\begin{array}{c}\frac{1}{{2n}} = \frac{1}{{2\left( {17} \right)}}\\ = 0.0294\end{array}\)
0	\(\begin{array}{c}\frac{3}{{2n}} = \frac{3}{{2\left( {17} \right)}}\\ = 0.0882\end{array}\)
0	\(\begin{array}{c}\frac{5}{{2n}} = \frac{5}{{2\left( {17} \right)}}\\ = 0.1471\end{array}\)
0	\(\begin{array}{c}\frac{7}{{2n}} = \frac{7}{{2\left( {17} \right)}}\\ = 0.2059\end{array}\)
0	\(\begin{array}{c}\frac{9}{{2n}} = \frac{9}{{2\left( {17} \right)}}\\ = 0.2647\end{array}\)
0	\(\begin{array}{c}\frac{{11}}{{2n}} = \frac{{11}}{{2\left( {17} \right)}}\\ = 0.3235\end{array}\)
1	\(\begin{array}{c}\frac{{13}}{{2n}} = \frac{{13}}{{2\left( {17} \right)}}\\ = 0.3824\end{array}\)
5	\(\begin{array}{c}\frac{{15}}{{2n}} = \frac{{15}}{{2\left( {17} \right)}}\\ = 0.4412\end{array}\)
6	\(\begin{array}{c}\frac{{17}}{{2n}} = \frac{{17}}{{2\left( {17} \right)}}\\ = 0.5000\end{array}\)
17	\(\begin{array}{c}\frac{{19}}{{2n}} = \frac{{19}}{{2\left( {17} \right)}}\\ = 0.5588\end{array}\)
24	\(\begin{array}{c}\frac{{21}}{{2n}} = \frac{{21}}{{2\left( {17} \right)}}\\ = 0.6176\end{array}\)
55	\(\begin{array}{c}\frac{{23}}{{2n}} = \frac{{23}}{{2\left( {17} \right)}}\\ = 0.6765\end{array}\)
91	\(\begin{array}{c}\frac{{25}}{{2n}} = \frac{{25}}{{2\left( {17} \right)}}\\ = 0.7353\end{array}\)
117	\(\begin{array}{c}\frac{{27}}{{2n}} = \frac{{27}}{{2\left( {17} \right)}}\\ = 0.7941\end{array}\)
155	\(\begin{array}{c}\frac{{29}}{{2n}} = \frac{{29}}{{2\left( {17} \right)}}\\ = 0.8529\end{array}\)
162	\(\begin{array}{c}\frac{{31}}{{2n}} = \frac{{31}}{{2\left( {17} \right)}}\\ = 0.9118\end{array}\)
205	\(\begin{array}{c}\frac{{33}}{{2n}} = \frac{{33}}{{2\left( {17} \right)}}\\ = 0.9706\end{array}\)

The corresponding z-scores of the areas computed above are tabulated below:

Areas	z-scores
0.0294	-1.890
0.0882	-1.352
0.1471	-1.049
0.2059	-0.821
0.2647	-0.629
0.3235	-0.458
0.3824	-0.299
0.4412	-0.148
0.5000	0.000
0.5588	0.148
0.6176	0.299
0.6765	0.458
0.7353	0.629
0.7941	0.821
0.8529	1.049
0.9118	1.352
0.9706	1.890

Now, plot the original data values on the x-axis and the corresponding z-scores on the y-axis.

Sample Values (x)	z-scores (y)
0	-1.890
0	-1.352
0	-1.049
0	-0.821
0	-0.629
0	-0.458
1	-0.299
5	-0.148
6	0.000
17	0.148
24	0.299
55	0.458
91	0.629
117	0.821
155	1.049
162	1.352
205	1.890

• Mark the values 0, 50, 100, ……., 250 on the horizontal scale. Label the axis as “Sample Value”.
• Mark the values -2.500, -2.000, -1.500, …….., 2.500 on the vertical axis. Label the axis as “z-score”.
• Place a dot for the values of the z-scores corresponding to the sample values on the x-axis.
• Draw a straight trend line.

The following normal quantile plot is obtained:

Assessing the normality of the values

It can be seen that the points on the plot do not follow a straight-line pattern.

Therefore, the given sample of the duration of tobacco use in non-Disney movies does not appear to come from a normally distributed population.

The t distribution (used for conducting the hypothesis test) is robust against departures from normality.This means that if the population of the sample does not appear to be normally distributed and the sample size is small, the test will still give accurate results and can be used for concluding the claim.

Thus, the non-normality of the 17 non-Disney times does not significantly affect the results regarding the claim.