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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

Calculate Sample Means

First, find the sample mean for both cable television and radio. Add all the hours spent watching television and divide by the number of individuals (15). Do the same for radio.For cable television: \\[\bar{X}_{TV} = \frac{22 + 8 + 25 + 22 + 12 + 26 + 22 + 19 + 21 + 23 + 14 + 14 + 14 + 16 + 24}{15} = 18.93\ \]For radio:\[\bar{X}_{Radio} = \frac{25 + 10 + 29 + 19 + 13 + 28 + 23 + 21 + 21 + 23 + 15 + 18 + 17 + 15 + 23}{15} = 19.87\]

02

Calculate Sample Variances

Next, calculate the sample variance for both groups. Variance measures how much the data points differ from the mean.For cable television:\[S^2_{TV} = \frac{\sum (X_i - \bar{X}_{TV})^2}{n-1} = \frac{\sum (X_i - 18.93)^2}{14} = 33.38\]For radio:\[S^2_{Radio} = \frac{\sum (X_i - \bar{X}_{Radio})^2}{n-1} = \frac{\sum (X_i - 19.87)^2}{14} = 42.84\]

03

Conduct t-test for Mean Difference

Determine if there is a significant difference between the means using a two-sample t-test. The null hypothesis is that the means are equal versus the alternative hypothesis that they are not equal.Calculate the test statistic:\[t = \frac{\bar{X}_{TV} - \bar{X}_{Radio}}{\sqrt{\frac{S^2_{TV}}{n} + \frac{S^2_{Radio}}{n}}} = \frac{18.93 - 19.87}{\sqrt{\frac{33.38}{15} + \frac{42.84}{15}}} = -0.39\]With 28 degrees of freedom, find the critical t-value and compare it to the calculated t-value. Also, find the p-value.

04

Compare p-value to Significance Level

The p-value is determined from the t-distribution table based on the calculated t-statistic and degrees of freedom. With a t-value of -0.39 and a 0.05 significance level, the p-value is approximately 0.702, which is greater than 0.05. Since the p-value is greater than 0.05, we do not reject the null hypothesis, indicating no significant difference in mean usage.

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Key Concepts

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

Sample Mean

The sample mean is a measure of the central tendency of a dataset. It is calculated by summing all individual data points and dividing by the total number of data points in the sample. This value gives us an idea of what a typical data point looks like. In the exercise, researchers calculated the sample mean for both cable television and radio usage.

• For cable television, the sample mean was calculated as \( \bar{X}_{TV} = 18.93 \).
• For radio, it was \( \bar{X}_{Radio} = 19.87 \).

This tells us that, on average, participants tend to spend slightly more time listening to the radio than watching cable television during the week.By looking at these averages, researchers can better understand media consumption patterns in the sample.

Sample Variance

Sample variance is a measure of the spread or dispersion of a set of data points in relation to the sample mean. It indicates how much the individual data points differ from the sample mean, which is crucial for understanding variability.To compute the sample variance, we subtract the sample mean from each data point, square each result, and then average these squared differences. The sample variance for:

• Cable television usage was calculated as \( S^2_{TV} = 33.38 \).
• Radio usage was \( S^2_{Radio} = 42.84 \).

A higher variance implies that the data points are more spread out from the mean, indicating more variability. In this case, radio usage displays a higher variance compared to cable television. This suggests there is more inconsistency in the number of hours individuals spend listening to the radio weekly compared to watching cable TV.

T-test

The T-test is a statistical test used to determine if there is a significant difference between the means of two groups. In this exercise, a two-sample T-test was used to compare the mean weekly usage of cable television and radio.

• The null hypothesis (H_0) states there is no difference between the population means for cable television and radio.
• The alternative hypothesis (H_1) proposes that there is a difference.

To perform the T-test, we use the formula:\[t = \frac{\bar{X}_{TV} - \bar{X}_{Radio}}{\sqrt{\frac{S^2_{TV}}{n} + \frac{S^2_{Radio}}{n}}}\]In this case, the calculated t-statistic was \(-0.39\). By comparing this with critical values from the t-distribution, we can evaluate if any observed difference is statistically significant.

P-value

The p-value is vital in hypothesis testing as it helps determine the significance of the results. It indicates the probability of observing the effect of data, assuming the null hypothesis is true. In hypothesis testing, you compare the p-value against a pre-determined significance level (alpha), usually set at 0.05. If the p-value is less than the significance level, you reject the null hypothesis, indicating a statistically significant result. In this exercise:

• The p-value was found to be approximately 0.702.
• Since 0.702 > 0.05, we do not reject the null hypothesis, suggesting no significant difference in mean media usage between cable television and radio.

Understanding the p-value gives researchers confidence in their conclusions regarding the observed data and hypothesized relationships.