Assume that the two samples are independent simple random...

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 $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Bad Stuff in Children's Movies Data Set 11 "Alcohol and Tobacco in Movies" in Appendix B includes lengths of times (seconds) of tobaccouse shown in animated children's movies. For the Disney movies, $n=33, \bar{x}=61.6 \mathrm{sec}, s=118.8$ sec. For the other movies, $n=17$ $\bar{x}=49.3 \mathrm{sec}, s=69.3 \mathrm{sec} .$ 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? $$\begin{array}{llllllllllllllll} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 & 6 & 17 & 24 & 55 & 91 & 117 & 155 & 162 & 205 \end{array}$$

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 $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ )
Bad Stuff in Children's Movies Data Set 11 "Alcohol and Tobacco in Movies" in Appendix B includes lengths of times (seconds) of tobaccouse shown in animated children's movies. For the Disney movies, $n=33, \bar{x}=61.6 \mathrm{sec}, s=118.8$ sec. For the other movies, $n=17$ $\bar{x}=49.3 \mathrm{sec}, s=69.3 \mathrm{sec} .$ 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?
$$\begin{array}{llllllllllllllll}
0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 & 6 & 17 & 24 & 55 & 91 & 117 & 155 & 162 & 205
\end{array}$$

Key Concepts

Two-Sample t-Test

This test compares the means of two independent groups to determine if there is a statistically significant difference between them. It is used when each sample is randomly selected from normally distributed populations, and the procedure can accommodate unequal variances through modifications like Welch's t-test.

Welch's t-Test

Welch's t-test is an adaptation of the traditional two-sample t-test that does not assume equal variances between the two populations. It adjusts the degrees of freedom based on the sample variances and sizes, making it more reliable when the assumption of homogeneity of variances is violated.

Null Hypothesis Significance Testing

This framework involves formulating a null hypothesis (typically stating that there is no difference between population means) and an alternative hypothesis, then using sample data to decide whether to reject the null in favor of the alternative. The decision is based on a predetermined significance level, which quantifies the risk of committing a Type I error.

Confidence Interval Construction

Confidence intervals provide a range of plausible values for the difference between population means and are constructed based on sample statistics, the chosen significance level, and appropriate degrees of freedom. They offer a dual perspective to hypothesis testing and indicate the precision of the estimated difference.

Normality Assumption

The assumption that the data are drawn from normally distributed populations is fundamental to the validity of the t-tests. When sample sizes are small, deviations from normality can significantly affect the test results, while larger samples tend to be more robust due to the Central Limit Theorem.

Degrees of Freedom

Degrees of freedom are used to determine the appropriate t-distribution for hypothesis testing and confidence interval construction. In tests where variances are unequal, methods like the Welch-Satterthwaite approximation yield a more accurate (often non-integer) degrees of freedom than simply taking the smaller of the two sample sizes minus one.

Transcript

00:01 So we would be assuming that the mean number of minutes for disney movies and the mean number of minutes of non -disney or other movies, children's movies, showing tobacco that those two are equal.

00:14 And alternately, that they're just different.

00:17 So again, we're looking at how many minutes or how many seconds, actually, we're seeing tobacco shown.

00:23 So we're going to assume that the difference between these two groups is zero.

00:31 And we're doing a two -tail test.

00:33 And so let's find what our test statistics is going to be.

00:37 And our smaller sample size is 17.

00:40 So we'll say we have 16 degrees of freedom.

00:43 And so i take the difference, 61 .6 seconds for disney, minus 49 .3 seconds for the other children movies.

00:52 And then we have very, very large standard deviations.

00:56 And that is something to really pay attention to.

00:59 Notice that those standard deviations look how much larger they are than the actual even mean.

01:06 And so normally, if you look at these two times, you're going to think, well, they're going to be significantly different because, you know, they're over 10 seconds apart.

01:15 However, with those huge standard deviations, that's going to make that be suspicious.

01:21 And when we get that, we get 0 .46.

01:25 So we're getting a test statistic here that is a.

01:29 T value that is 0 .46 and this area plus going symmetrically on the other side those two together are going to be our p value so we need to find two times the probability of that being greater than or equal to 0 .46 and i used my tcdf on my calculator and found out that that p value is about 65 so if you have two distributions that their difference is actually zero, the likelihood of getting what we got or more extreme in each direction happens 65 % of the time.

02:06 And this is definitely not less than 5%.

02:10 Therefore, we fail to reject null.

02:18 So it appears as though the means are equal.

02:21 So they mean amount of time that you see tobacco means are equal...

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