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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.

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

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

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

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

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.

\({s_2} = 69.3\)

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.

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.

a.

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

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

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.

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).

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:

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

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

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

• Mark the values 0, 50, 100, 鈥︹��., 250 on the horizontal scale. Label the axis as 鈥淪ample Value鈥�.
• Mark the values -2.500, -2.000, -1.500, 鈥︹��.., 2.500 on the vertical axis. Label the axis as 鈥渮-score鈥�.
• 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:

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.