91Ó°ÊÓ

In recent years, a growing array of entertainment options competes for consumer time. By \(2004,\) cable television and radio surpassed broadcast television, recorded music, and the daily newspaper to become the two entertainment media with the greatest usage (The Wall Street Journal, January 26,2004 ). Researchers used a sample of 15 individuals and collected data on the hours per week spent watching cable television and hours per week spent listening to the radio. $$\begin{array}{ccccc} \text { Individual } & \text { Television } & \text { Radio } & \text { Individual } & \text { Television } & \text { Radio } \\ 1 & 22 & 25 & 9 & 21 & 21 \\ 2 & 8 & 10 & 10 & 23 & 23 \\ 3 & 25 & 29 & 11 & 14 & 15 \\ 4 & 22 & 19 & 12 & 14 & 18 \\ 5 & 12 & 13 & 13 & 14 & 17 \\ 6 & 26 & 28 & 14 & 16 & 15 \\ 7 & 22 & 23 & 15 & 24 & 23 \\ 8 & 19 & 21 & & \end{array}$$ a. Use a .05 level of significance and test for a difference between the population mean usage for cable television and radio. What is the \(p\) -value? b. What is the sample mean number of hours per week spent watching cable television? What is the sample mean number of hours per week spent listening to radio? Which medium has the greater usage?

Step by step solution

01

Define the Hypotheses

To test for a difference in the population mean usage for cable television and radio, we set up the following null and alternative hypotheses: - Null Hypothesis (H0): \( \mu_{TV} = \mu_{Radio} \), meaning there is no difference in mean usage between the two media.- Alternative Hypothesis (H1): \( \mu_{TV} eq \mu_{Radio} \), meaning there is a difference in mean usage between the two media.

02

Calculate Sample Means and Differences

Calculate the sample means for television and radio. Then, find the differences for each individual.For Television:- Total hours for Television \((\sum = 266)\)- Sample mean for Television \(\left(\bar{x}_{TV} = \frac{266}{15} = 17.73\right)\)For Radio:- Total hours for Radio \((\sum = 280)\)- Sample mean for Radio \(\left(\bar{x}_{Radio} = \frac{280}{15} = 18.67\right)\)The differences are calculated for each individual as: \( TV_i - Radio_i \).

03

Calculate the Test Statistic

The test statistic for a paired t-test is computed using the differences:- Differences: \([-3, -2, -4, 3, -1, -2, -1, -2, 0, 0, -1, -4, -3, 1, 1]\)- Mean of differences \(\bar{d} = \frac{-23}{15} = -1.53\)- Sum of squared differences (SSD) = \(120\)- Standard deviation of differences \(s_d = \sqrt{\frac{1}{n-1} \cdot SSD} \approx 2.42\)Compute the t-statistic as:\[ t = \frac{\bar{d}}{\frac{s_d}{\sqrt{n}}} = \frac{-1.53}{\frac{2.42}{\sqrt{15}}} \approx -2.06.\]

04

Determine the p-value

Using a t-distribution table or calculator, find the p-value for \(t = -2.06\) with \(n-1 = 14\) degrees of freedom. The p-value is approximately \(0.059\).

05

Make a Decision

Compare the p-value to the level of significance:- The p-value \(0.059\) is greater than the significance level \(0.05\).Since \(p > \alpha\), we fail to reject the null hypothesis. There is not enough evidence to conclude a difference in mean usage between cable TV and radio.

06

Conclusion

For part (a), there is no significant difference between the means of television and radio usage at a 0.05 level of significance. The p-value is approximately 0.059. For part (b), the average weekly hours for television and radio are 17.73 and 18.67, respectively. Radio has slightly greater usage than television.

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Key Concepts

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

Paired t-test

The paired t-test is a statistical method used to compare two related samples. It's particularly useful when you want to examine differences between paired observations, such as measurements taken from the same subjects under different conditions. In our exercise, the paired t-test helps us discover whether there is a significant difference in media consumption between hours spent watching cable television and listening to the radio.

Before you calculate the test statistic, the first step is to determine the null and alternative hypotheses:

• Null Hypothesis (\( H_0 \)): There is no difference in means, meaning that the population mean of TV usage is equal to the population mean of radio usage (\( \mu_{TV} = \mu_{Radio} \)).
• Alternative Hypothesis (\( H_1 \)): There is a difference in means (\( \mu_{TV} eq \mu_{Radio} \)).

Next, calculate the differences for each individual between TV and radio hours, and then find their mean. Ultimately, this paired t-test is about evaluating if the observed mean of differences is statistically significant.

Population Mean

The population mean is the average value of a given variable in an entire population. It's denoted by the Greek letter \( \mu \), and in research, it's often an unknown quantity you're trying to estimate.

In hypothesis testing, such as the paired t-test, we aim to compare the population means of two related groups based on our sample data. Here, we're exploring whether the population mean for hours spent on cable television differs from that of radio. Although the exercise uses sample data (15 individuals) to test our hypotheses, the goal is to infer something about these larger population averages.

When you conduct this test yourself, you'll use sample means to draw conclusions about these population means. **It's important to remember that sample data is a representation of our larger population and helps us make educated estimations of population means.**

Sample Mean

The sample mean is the average of a data sample, a subset of your total population. It's represented by \( \overline{x} \) and is a critical component in hypothesis testing, providing a basis for comparing both groups in your study.

In this exercise, we calculated:

• Sample Mean for Television: \( \overline{x}_{TV} = 17.73 \), computed from the total hours divided by the number of individuals.
• Sample Mean for Radio: \( \overline{x}_{Radio} = 18.67 \), calculated in the same way as television hours.

These sample means offer insight into how much time individuals in our sample allocate to each media form. By comparing them, you can check for variances and make inferences about how the average population might behave. Recognize that sample means are an approximation of your respective population means and inherently come with a degree of sampling error.

Significance Level

The significance level, denoted as \( \alpha \), is a threshold used in hypothesis testing that dictates when we should reject the null hypothesis. In practical terms, it defines how willing you are to tolerate a false positive—a situation where you incorrectly conclude there's an effect when there isn't one.

In this problem, the significance level is set at 0.05, or 5%. This threshold suggests that if the p-value you calculate is below 0.05, you have enough evidence to reject the null hypothesis. In contrast, a p-value greater than 0.05 does not provide that evidence, so you fail to reject the null.

Applying the significance level ensures our decisions are statistically justified, thereby reducing the likelihood of misinterpretation. In the given exercise, the computed p-value was 0.059, which did not meet our significance level, leading us to conclude that there was not a significant difference between the two media usage rates. Always be mindful of setting your significance level before testing as it guides your decision-making process.