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Chapter 15: Problem 47

In a particular metro area, there are three commercial television stations, each with its own news program from 6: 00 to 6: 30 p.m. According to a report in this morning's local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11 ), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watching the three channels? \(?\) ? hannels?

Short Answer

Expert verified
Yes, there is a significant difference in the proportion of viewers watching the three channels.

Step by step solution

01

Define the Hypotheses

We will perform a Chi-Square test for goodness of fit. The null hypothesis, denoted as \( H_0 \), is that there is no difference in the proportion of viewers watching the three channels. The alternative hypothesis, \( H_a \), is that there is a difference in the proportion of viewers watching the channels.
02

Calculate Expected Frequencies

The total number of viewers sampled is 150. Under the null hypothesis, each channel would be expected to have an equal number of viewers. Hence, the expected frequency for each channel is \( \frac{150}{3} = 50 \).
03

Determine Observed Frequencies

The observed frequencies are as follows: 53 viewers for WNAE, 64 for WRRN, and 33 for WSPD.
04

Calculate the Chi-Square Statistic

Use the formula: \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. Calculate for each channel: - WNAE: \( \frac{(53-50)^2}{50} = 0.18 \) - WRRN: \( \frac{(64-50)^2}{50} = 3.92 \) - WSPD: \( \frac{(33-50)^2}{50} = 5.78 \) Thus, \( \chi^2 = 0.18 + 3.92 + 5.78 = 9.88 \).
05

Determine Critical Value and Decision

The degrees of freedom \( df \) for this test is \( k-1 = 2 \) where \( k \) is the number of categories (channels). At a significance level of 0.05, the critical value from the Chi-Square distribution table for 2 degrees of freedom is approximately 5.991. Since our calculated \( \chi^2 \) value of 9.88 is greater than 5.991, we reject the null hypothesis.

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Key Concepts

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

Goodness of Fit
In statistics, the goodness of fit test helps us understand how well our observed data matches the expected data according to a specific hypothesis. It鈥檚 particularly useful when you have categorical data and want to see if certain counts follow a particular distribution.
The exercise here utilizes a Chi-Square test for goodness of fit. This type of test is perfect for checking whether the observed frequencies of viewers for each TV channel fit well with what we would expect them to be if each channel had an equal viewership share. Basically, it helps us check if there are significant differences in how many people watched each channel.
Hypothesis Testing
Hypothesis testing is a method used to decide whether there's enough evidence to reject a default assumption, known as the null hypothesis. In the context of this Chi-Square test, the null hypothesis ( H_0 ) states that there is no difference in the proportions of viewers among the three TV channels. Meanwhile, the alternative hypothesis ( H_a ) suggests that some difference does exist.
The process typically involves comparing the observed data to what we'd anticipate under the null hypothesis. If the data deviates significantly from what we'd expect, we might reject the null hypothesis in favor of the alternative hypothesis.
Significance Level
The significance level, denoted as \(\alpha\), is a crucial part of hypothesis testing. It essentially represents the threshold at which we decide whether to reject the null hypothesis. By convention, a significance level of 0.05 is often used.
This means that if our test statistic falls in the top 5% of all possible values it could take under the null hypothesis, we reject the null hypothesis. For this exercise, the significance level is 0.05, indicating a 5% risk of concluding that a difference exists when it doesn't.
Degrees of Freedom
Degrees of freedom in statistical tests provide insight into the number of categories or variables that can vary freely while estimating others. For a Chi-Square test for goodness of fit, the formula for calculating degrees of freedom is the number of categories minus one: \(df = k - 1\).
In our scenario, since there are three television channels, there are 2 degrees of freedom (3 - 1 = 2). This impacts which Chi-Square distribution we compare our calculated statistic against to decide if we should reject the null hypothesis.

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Most popular questions from this chapter

A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Do these data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?

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