Problem 12 - Basic Statistics for Business and Economics with Student

Question

For many years TV executives used the guideline that 30 percent of the audience were watching each of the traditional big three prime-time networks and 10 percent were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 165 homes were tuned in to the \(\mathrm{ABC}\) affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the .05 significance level, can we conclude that the guideline is still reasonable?

Step-by-Step Solution

Answer

The guideline is not reasonable based on the observed data.

Step 1: Formulate Hypotheses

Set up the null and alternative hypotheses. - Null Hypothesis \(H_0\): The distribution of viewers follows the guideline.- Alternative Hypothesis \(H_a\): The distribution of viewers does not follow the guideline.

Step 2: Calculate Expected Frequencies

The total number of viewers is 500. According to the guideline, 30% should watch each network (ABC, CBS, NBC), and 10% should watch cable. Thus, Expected frequency for ABC = 0.3 × 500 = 150, Expected frequency for CBS = 0.3 × 500 = 150, Expected frequency for NBC = 0.3 × 500 = 150, Expected frequency for Cable = 0.1 × 500 = 50.

Step 3: Calculate Chi-Squared Statistic

Use the chi-squared statistic 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.Calculated values:- \(O_{\text{ABC}} = 165, E_{\text{ABC}} = 150\);- \(O_{\text{CBS}} = 140, E_{\text{CBS}} = 150\);- \(O_{\text{NBC}} = 125, E_{\text{NBC}} = 150\);- \(O_{\text{Cable}} = 70 (500 - 165 - 140 - 125), E_{\text{Cable}} = 50\).Calculate:\[\chi^2 = \frac{(165-150)^2}{150} + \frac{(140-150)^2}{150} + \frac{(125-150)^2}{150} + \frac{(70-50)^2}{50}\]\[\chi^2 = \frac{225}{150} + \frac{100}{150} + \frac{625}{150} + \frac{400}{50}\]\[\chi^2 = 1.5 + 0.67 + 4.17 + 8\]\[\chi^2 = 14.34\].

Step 4: Determine Critical Value

The degrees of freedom \(df\) is the number of categories minus 1: \(df = 4 - 1 = 3\). For \(\alpha = 0.05\) and \(df = 3\), the critical value from the chi-squared distribution table is 7.815.

Step 5: Compare Chi-Squared Statistic to Critical Value

Since the calculated \(\chi^2 = 14.34\) is greater than the critical value 7.815, we reject the null hypothesis.

Step 6: Conclusion of Hypothesis Test

Since we rejected the null hypothesis, there is enough evidence at the 0.05 significance level to conclude that the viewing distribution does not follow the guideline.

Key Concepts

Chi-Squared Test

A chi-squared test, in particular the chi-squared goodness of fit test, helps us determine how well our observed data matches expectations based on an assumed model or guideline. Essentially, it checks if there's a significant difference between what we collect as data and what was proposed or expected.
In the given problem regarding TV viewership, the chi-squared test assesses whether the observed distribution of TV viewers—those watching ABC, CBS, NBC, and cable—aligns with an expected distribution, according to an older guideline.
To apply this test, we calculate the chi-squared statistic with this 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.
Once the chi-squared statistic is calculated, it is compared to a critical value from the chi-squared distribution table, based on the degrees of freedom and chosen significance level.

Significance Level

The significance level, denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It's a threshold for decision-making in hypothesis testing, giving us a benchmark for how much evidence we require to support our conclusion.
Typically, a common choice for \( \alpha \) is 0.05, which implies a 5% risk of concluding that a difference or effect exists when it doesn’t.
In our case with TV viewers, the significance level of 0.05 was used. This means there must be sufficient evidence to conclude that the current viewership distribution significantly deviate from the past guidelines, with only a 5% chance that this conclusion is due to random sample variation.
By setting this boundary, the hypothesis test becomes a more objective tool for decision-making. If the calculated chi-squared statistic surpasses the critical value at the 0.05 significance level, we've gathered strong enough evidence to claim that the viewership distribution does not adhere to the guidelines.

Null Hypothesis

The null hypothesis, abbreviated as \( H_0 \), is a statement that there's no effect or no difference, and it forms the basis for traditional hypothesis testing. It's the default or "no change" position that we aim to check against observable data.
In the TV viewership problem, the null hypothesis posits that the distribution of viewers across different networks follows the traditional guideline. Specifically, this implies that 30% of viewers watch each of the big three networks and 10% watch cable stations.
The purpose of the null hypothesis is to provide a clear statement that can be tested statistically. If the data significantly deviates from what the null hypothesis suggests, we may seek an alternative explanation for our observations.

Alternative Hypothesis

The alternative hypothesis, denoted by \( H_a \) or sometimes \( H_1 \), offers a competing theory—that there is a notable effect or difference in the data set that necessitates a different explanation from what the null hypothesis provides.
For the TV viewership example, the alternative hypothesis is that the distribution does not adhere to the old guidelines. This implies that people's viewing habits may have changed, and the percentages of the audience dedicated to each of the networks or cable are different now compared to the past.
Unlike the null hypothesis, the alternative hypothesis is often what researchers are most interested in confirming. If our data analysis leads to rejecting the null hypothesis, we provide support for the alternative, suggesting that a significant shift in viewer distribution has occurred.