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Chapter 9: Problem 11

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 \(\alpha=0.01\) $$\begin{array}{lll} & \text { Children } & \text { Teens } \\\\\hline \text { Sample mean } & 22.45 & 18.50 \\\\\text { Sample variance } & 16.4 & 18.2 \\\\\text { Sample size } & 15 & 15\end{array}$$

Short Answer

Expert verified
There is no significant evidence of a difference between the two groups' watching times at \( \alpha = 0.01 \).

Step by step solution

01

Define Hypotheses

We need to determine whether there is a significant difference in average watching times between children and teens. We set up our null hypothesis as \( H_0: \mu_1 = \mu_2 \), indicating no difference in means, and our alternative hypothesis as \( H_a: \mu_1 eq \mu_2 \), indicating a difference in watching times.
02

Determine the Test Statistic

We'll use a two-sample t-test for the means since the sample sizes are small. The test statistic \( t \) is calculated as: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] 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

Calculate the Standard Error

Calculate the standard error of the difference in means: \[ \text{SE} = \sqrt{\frac{16.4}{15} + \frac{18.2}{15}} \approx \sqrt{1.0933 + 1.2133} \approx 1.4897 \]
04

Compute the t-Statistic

Substitute the values into the formula for \( t \): \[ t = \frac{22.45 - 18.50}{1.4897} \approx \frac{3.95}{1.4897} \approx 2.652 \]
05

Define the Critical Region

With \( \alpha = 0.01 \) and degrees of freedom \( df = 14 + 14 - 2 = 28 \), we use a t-table to find a critical value for the two-tailed test. The critical t-value is approximately 2.763. Our test is at \( 0.01 \) significance level, so if \( |t| > 2.763 \), we reject \( H_0 \).
06

Make a Decision

Compare the calculated t-statistic to the critical value. Since \( 2.652 < 2.763 \), we do not reject the null hypothesis. This means there is insufficient evidence to claim a difference in average television watching times between children and teens at the 0.01 significance level.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics, allowing us to decide if data supports a specific claim or not. In the context of comparing how much TV children and teens watch, we form two hypotheses. The **null hypothesis** \(H_0\) states that there is no difference in the average viewing times between the two groups \((\mu_1 = \mu_2)\). The **alternative hypothesis** \(H_a\) suggests there is a difference \((\mu_1 eq \mu_2)\).
  • **Null Hypothesis \(H_0\):** Asserts no effect or relationship; in this case, it says children and teens watch the same amount of TV on average.
  • **Alternative Hypothesis \(H_a\):** Contradicts \(H_0\); claims there is a measurable difference in TV watching habits.
  • Hypothesis testing guides decision-making by providing a structured method to determine which hypothesis is supported by the data.
The hypotheses form the foundation from which we decide if the observed data merely reflects random variation or a real effect.
Sample Statistics
Sample statistics are key in dedicating what hypotheses are worth investigating further. They include information like means and variances derived from our samples. Here, we measure:
  • **Sample Means:** For children \((\bar{x}_1 = 22.45)\) and teens \((\bar{x}_2 = 18.50)\), they represent the average TV hours per week each group watches.
  • **Sample Variances:** For children \((s_1^2 = 16.4)\) and teens \((s_2^2 = 18.2)\), they reflect the variability or spread of TV hours in each group.
  • **Sample Sizes:** Both groups consist of 15 individuals \((n_1 = n_2 = 15)\).
The analysis begins with these descriptive statistics, allowing us to quantify the behavior of each group and make informed decisions using inferential statistics techniques.
Standard Error
The standard error helps us to understand the precision of our sampled data statistics by gauging the variability between sample means. When we conduct a two-sample t-test, the standard error \(SE\) of the difference in means allows us to consider whether any observed changes between our groups might easily occur by random chance.
We calculate the standard error using:\[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
  • Here, \(SE \approx 1.4897\) quantifies how much the sample means would fluctuate if we repeatedly sampled from the population under identical conditions.
  • A smaller \(SE\) suggests greater reliability in our results indicating that the sample mean is a good estimator of the population mean; conversely, a larger \(SE\) points to less reliability.
Thus, standard error is crucial in assessing the significance of our results by influencing the t-statistic.
Critical Value
The critical value is a threshold we use to decide whether the test statistic indicates a significant departure from the null hypothesis. It is determined by the chosen level of significance \((\alpha)\) and the degrees of freedom attributed to the test.
  • **Significance Level \((\alpha)\):** Here, 0.01 means we're screening for strong evidence to reject \(H_0\).
  • **Degrees of Freedom:** Calculated as the total number of samples minus two; in this case, it's 28.
  • **Critical Value:** Based on a two-tailed test and the given \(\alpha\), we determine the critical t-value to be approximately 2.763 from a t-distribution table or calculator.
The critical value frames our decision-making: if the absolute value of our observed t-statistic surpasses this threshold, we reject the null hypothesis. If not, as we've seen with \(2.652 < 2.763\), we conclude there's not enough statistical evidence to affirm a difference in TV habits at the prescribed significance level.

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Most popular questions from this chapter

Perform each of the following 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. A random survey of 1000 students nationwide showed a mean ACT score of 21.4. Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is 3 , can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

Find each \(X\), given \(\hat{p}\). a. \(\hat{p}=0.24, n=300\) b. \(\hat{p}=0.09, n=200\) c. \(\hat{p}=88 \%, n=500\) d. \(\hat{p}=40 \%, n=480\) e. \(\hat{p}=32 \%, n=700\)

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. The weights in ounces of a random sample of running shoes for men and women are shown. Calculate the variances for each sample, and test the claim that the variances are equal at \(\alpha=0.05\). Use the \(P\) -value method. $$ \begin{array}{rrr|rrr} && {\text { Men }} & {\text { Women }} \\ \hline 11.9 & 10.4 & 12.6 & 10.6 & 10.2 & 8.8 \\ 12.3 & 11.1 & 14.7 & 9.6 & 9.5 & 9.5 \\ 9.2 & 10.8 & 12.9 & 10.1 & 11.2 & 9.3 \\ 11.2 & 11.7 & 13.3 & 9.4 & 10.3 & 9.5 \\ 13.8 & 12.8 & 14.5 & 9.8 & 10.3 & 11.0 \end{array} $$

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. In a random sample of 80 Americans, 44 wished that they were rich. In a random sample of 90 Europeans, 41 wished that they were rich. At \(\alpha=0.01,\) is there a difference in the proportions? Find the \(99 \%\) confidence interval for the difference of the two proportions.

Perform each of these steps. Assume that all variables are normally or approximately normally distributed 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. Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were randomly selected. The rates are given in the table as number of beats per minute. At \(\alpha=0.01,\) is there a significant difference in the average pulse rates of twins? Use the \(P\) -value method. Find the \(99 \%\) confidence interval for the difference of the two. $$ \begin{array}{l|llllllll} \text { Twin A } & 87 & 92 & 78 & 83 & 88 & 90 & 84 & 93 \\ \hline \text { Twin B } & 83 & 95 & 79 & 83 & 86 & 93 & 80 & 86 \end{array} $$
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