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Chapter 9: Problem 11

Hours Spent Watching Television According to Nielsen Media Research, children (ages 2-11) spend an average of 21 hours 30 minutes watching television per week while teens (ages 12-17) spend an average of 20 hours 40 minutes. Based on the sample statistics shown, is there sufficient evidence to conclude a difference in average television watching times between the two groups? Use \(\alpha=0.01 .\) $$ \begin{array}{lcc}{} & {\text { Children }} & {\text { Teens }} \\ \hline \text { Sample mean } & {22.45} & {18.50} \\ {\text { Sample variance }} & {16.4} & {18.2} \\ {\text { Sample size }} & {15} & {15}\end{array} $$

Short Answer

Expert verified
No, there is not enough evidence to conclude a difference in average TV watching times at \(\alpha = 0.01\).

Step by step solution

01

State the Hypotheses

First, we need to establish the null hypothesis and the alternative hypothesis. The null hypothesis \(H_0\) states that there is no difference in the average television watching times between children and teens: \(H_0: \mu_1 = \mu_2\). The alternative hypothesis \(H_a\) states that there is a difference: \(H_a: \mu_1 eq \mu_2\).
02

Calculate the Test Statistic

We'll use the two-sample t-test formula for comparing the means of two independent groups. The formula for the t-statistic is: \[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]Where \(\bar{x}_1 = 22.45\), \(\bar{x}_2 = 18.50\), \(s_1^2 = 16.4\), \(s_2^2 = 18.2\), \(n_1 = 15\), \(n_2 = 15\).
03

Perform the Calculation

Substitute the values into the formula:\[t = \frac{22.45 - 18.50}{\sqrt{\frac{16.4}{15} + \frac{18.2}{15}}}\]Calculate the differences and the square root:\[t = \frac{3.95}{\sqrt{1.0933 + 1.2133}} = \frac{3.95}{1.471}\]\[t \approx 2.69\]
04

Determine the Critical Value and Compare

Given \(\alpha = 0.01\) and a two-tailed test, with \(n_1 + n_2 - 2 = 28\) degrees of freedom, we find the critical t-value using a t-distribution table. The critical value for \(\alpha = 0.01\) is approximately \(\pm 2.763\). Since our calculated \(t\) statistic (2.69) is less than 2.763, we do not reject the null hypothesis.
05

Conclusion

Since the calculated t-statistic is less than the critical value, there is not enough evidence to reject the null hypothesis. Therefore, we conclude that there is insufficient evidence to suggest a significant difference in average television watching times between children and teens at \(\alpha = 0.01\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-Sample t-Test
The two-sample t-test is a powerful statistical method used when we want to compare the means of two independent groups. In our example, it helps us determine if the average time spent watching television per week by children is significantly different from that of teens.

Three key assumptions in this test include:
  • The populations should have normal distributions.
  • The variances of the two populations are equal.
  • The samples are independent.
In practice, the two-sample t-test involves calculating a t-statistic, which tells us how far apart our sample means are in terms of standard errors. This is why the formula includes terms that account for sample variability and size. A larger absolute t-statistic indicates a bigger difference between the means.
Critical Value
The critical value in hypothesis testing is a cutoff point that determines the threshold for deciding whether to reject or not reject the null hypothesis. In the given exercise, the significance level, denoted by \( \alpha \), is set at 0.01, indicating a 1% risk of making a Type I error (rejecting a true null hypothesis).

The critical value is the boundary that separates extreme values from those that are considered normal. Since we are conducting a two-tailed test, we must check the t-distribution table to find the critical value for a 0.01 significance level and the appropriate degrees of freedom. Values beyond this critical point suggest significant evidence against the null hypothesis, prompting its rejection.
t-Distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but with thicker tails. It is especially useful when dealing with small sample sizes, as it provides a more accurate estimate of variability. The shape of the t-distribution depends on the degrees of freedom, which we'll discuss next.

This distribution becomes very useful when conducting a t-test because it allows us to assess the probability of obtaining a t-statistic as extreme as what we've calculated, under the assumption that the null hypothesis is true. As the sample size increases, the t-distribution becomes closer to a normal distribution.
Degrees of Freedom
Degrees of freedom (df) are a critical concept in statistics that refer to the number of values in a calculation that are free to vary. In the context of a two-sample t-test, the degrees of freedom help determine the shape of the t-distribution used in hypothesis testing.

For independent sample t-tests, df is calculated as the sum of the sample sizes minus two, which accounts for the estimation of two group means鈥攐ne for each group. For our exercise, with both groups having 15 samples each, the degrees of freedom are calculated as \(df = n_1 + n_2 - 2 = 28\). This calculation impacts the critical value we use to determine significance, affecting whether the difference between sample means can be considered statistically significant.

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Most popular questions from this chapter

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Married People In a specific year \(53.7 \%\) of men in the United States were married and \(50.3 \%\) of women were married. Two independent random samples of 300 men and 300 women found that 178 men and 139 women were married (not to each other). At the 0.05 level of significance, can it be concluded that the proportion of men who were married is greater than the proportion of women who were married?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. High School Graduation Rates The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\mathrm{a}\) difference of \(3 \%\). Random samples of 1200 students from each state indicated that 980 graduated in Pennsylvania and 940 graduated in Idaho. At the 0.05 level of significance, can it be concluded that there is a difference in the proportions of graduating students between the states?

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Test Scores An instructor who taught an online statistics course and a classroom course feels that the variance of the final exam scores for the students who took the online course is greater than the variance of the final exam scores of the students who took the classroom final exam. The following data were obtained. At \(\alpha=0.05\) is there enough evidence to support the claim? $$ \begin{array}{cc}{\text { Online Course }} & {\text { Classroom Course }} \\\ \hline s_{1=3.2} & {s_{2}=2.8} \\ {n_{1}=11} & {n_{2}=16}\end{array} $$

Sale Prices for Houses The average sales price of new one-family houses in the Midwest is dollar 250,000 and in the South is dollar 253,400. A random sample of 40 houses in each region was examined with the following results. At the 0.05 level of significance, can it be concluded that the difference in mean sales price for the two regions is greater than dollar 3400 ? $$ \begin{array}{lcc}{} & {\text { South }} & {\text { Midwest }} \\ \hline \text { Sample size } & {40} & {40} \\ {\text { Sample mean }} & {\$ 261,500} & {\$ 248,200} \\ {\text { Population standard deviation }} & {\$ 10,500} & {\$ 12,000}\end{array} $$

Average Earnings for College Graduates The average earnings of year-round full-time workers with bachelor's degrees or more is dollar 88,641 for men and dollar 58,000 for women - a difference of slightly over dollar 30,000 a year. One hundred of each were randomly sampled, resulting in a sample mean of dollar 90,200 for men, and the population standard deviation is dollar 15,000 ; and a mean of dollar 57,800 for women, and the population standard deviation is dollar 12,800. At the 0.01 level of significance, can it be concluded that the difference in means is not dollar 30,000 ?
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