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Chapter 15: Problem 18

The publisher of a sports magazine plans to offer new subscribers one of three gifts: a sweatshirt with the logo of their favorite team, a coffee cup with the logo of their favorite team, or a pair of earrings also with the logo of their favorite team. In a sample of 500 new subscribers, the number selecting each gift is reported below. At the .05 significance level, is there a preference for the gifts or should we conclude that the gifts are equally well liked? $$ \begin{array}{|lc|} \hline \text { Gift } & \text { Frequency } \\ \hline \text { Sweatshirt } & 183 \\ \text { Coffee cup } & 175 \\ \text { Earrings } & 142 \\ \hline \end{array} $$

Short Answer

Expert verified
No preference; gifts are equally liked.

Step by step solution

01

Define Hypotheses

We need to perform a Chi-square test to evaluate if there is a preference among gifts. Our null hypothesis \( H_0 \) is that there is no preference among gifts, meaning they are all equally liked. The alternative hypothesis \( H_a \) is that there is a preference for at least one of the gifts.
02

Calculate Expected Frequencies

If there is no preference, each gift should be selected equally among the 500 subscribers. Thus, the expected frequency for each gift is \( \frac{500}{3} \approx 166.67 \).
03

Compute Chi-square Test Statistic

The Chi-square statistic \( \chi^2 \) is calculated using: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. For Sweatshirts: \( \frac{(183 - 166.67)^2}{166.67} \approx 1.65 \); Coffee cups: \( \frac{(175 - 166.67)^2}{166.67} \approx 0.42 \); Earrings: \( \frac{(142 - 166.67)^2}{166.67} \approx 3.85 \). Summing these, \( \chi^2 \approx 5.92 \).
04

Determine Degrees of Freedom and Critical Value

The degrees of freedom \( df \) is equal to the number of categories minus one: \( df = 3 - 1 = 2 \). At a \( 0.05 \) significance level, the critical value can be found in a Chi-square distribution table and is approximately \( 5.991 \).
05

Make a Conclusion

Compare the calculated \( \chi^2 = 5.92 \) with the critical value \( 5.991 \). Since our test statistic is less than the critical value, we fail to reject the null hypothesis. This result indicates there is not enough evidence to suggest a preference among gifts.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a way to make decisions about a certain assumption or hypothesis using sample data. In this problem, we are using it to assess whether new magazine subscribers prefer one gift over the others.

There are two hypotheses:
  • **Null Hypothesis (\( H_0 \)):** This states that there is no difference in preference among the gifts. In other words, the gifts are equally liked.
  • **Alternative Hypothesis (\( H_a \)):** This suggests that at least one gift is preferred over the others.
When conducting hypothesis testing, we often try to provide evidence to reject the null hypothesis, thereby supporting the alternative hypothesis. In our case, if the result of our test shows no significant difference, we will fail to reject the null hypothesis, suggesting no preference among gifts.
Understanding Degrees of Freedom
The concept of degrees of freedom refers to the number of values in a statistical calculation that are free to vary.

In the context of the Chi-square test, the degrees of freedom help us determine the shape of the Chi-square distribution. The formula for degrees of freedom in this test is:
  • Degrees of freedom (\( df \)) = Number of categories - 1
Since we have three categories (sweatshirt, coffee cup, and earrings), the degrees of freedom would be:
  • \( df = 3 - 1 = 2 \)
Understanding degrees of freedom is essential as they determine the critical value in a Chi-square distribution table, which we then compare to our test statistic to make inferences about our hypotheses.
Understanding Significance Level
The significance level in hypothesis testing is the criterion used to decide whether a null hypothesis should be rejected. It is denoted by \( \alpha \)and reflects the probability of committing a Type I error. A Type I error occurs when we wrongly reject a true null hypothesis.

In this exercise, the significance level is set at 0.05, meaning there is a 5% risk of concluding that there is a preference among gifts when there actually isn't. When we compare our calculated test statistic against a critical value corresponding to this significance level, it determines our decision:
  • If the test statistic exceeds the critical value, we reject the null hypothesis.
  • If it does not exceed, we fail to reject the null hypothesis.
This approach allows us to control the likelihood of making incorrect inferences from our data.
Understanding Expected Frequency
Expected frequency in the Chi-square test refers to the frequency that we would expect to observe if there were no effect or association. It serves as a benchmark for comparison against the observed frequencies in each category.

To calculate expected frequency, we assume that each category should have the same number under the null hypothesis that all gifts are equally liked. For our gifts:
  • Total subscribers: 500
  • Number of categories: 3
  • Expected frequency for each gift: \( \frac{500}{3} \approx 166.67 \)
These expected frequencies are then used to compute the Chi-square statistic, comparing them against the observed frequencies. It helps determine if the discrepancies between what was observed and what was expected are significant or not.

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Most popular questions from this chapter

The use of cellular phones in automobiles has increased dramatically in the last few years. Of concern to traffic experts, as well as manufacturers of cellular phones, is the effect on accident rates. Is someone who is using a cellular phone more likely to be involved in a traffic accident? What is your conclusion from the following sample information? Use the .05 significance level. $$ \begin{array}{|lcc|} \hline & \begin{array}{c} \text { Had Accident } \\ \text { in the Last Year } \end{array} & \begin{array}{c} \text { Did Not Have an Accident } \\ \text { in the Last Year } \end{array} \\ \hline \text { Cellular phone in use } & 25 & 300 \\ \text { Cellular phone not in use } & 50 & 400 \\ \hline \end{array} $$

In a particular chi-square goodness-of-fit test there are six categories and 500 observations. Use the .01 significance level. a. How many degrees of freedom are there? b. What is the critical value of chi-square?

For many years TV executives used the guideline that 30 percent of the audience were watching each of the prime-time networks, that is \(\mathrm{ABC}, \mathrm{NBC}\) and \(\mathrm{CBS},\) 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 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?

In a particular chi-square goodness-of-fit test there are four categories and 200 observations. Use the .05 significance level. a. How many degrees of freedom are there? b. What is the critical value of chi-square? \(?\) ?

Two hundred managers employed in the lumber industry were randomly selected and interviewed regarding their concern about environmental issues. The response of each person was tallied into one of three categories: no concern, some concern, and great concern. The results were: $$ \begin{array}{|lccc|} \hline \begin{array}{l} \text { Level of } \\ \text { Management } \end{array} & \begin{array}{c} \text { No } \\ \text { Concern } \end{array} & \begin{array}{c} \text { Some } \\ \text { Concern } \end{array} & \begin{array}{c} \text { Great } \\ \text { Concern } \end{array} \\ \hline \text { Top management } & 15 & 13 & 12 \\ \text { Middle management } & 20 & 19 & 21 \\ \text { Supervisor } & 7 & 7 & 6 \\ \text { Group leader } & 28 & 21 & 31 \\ \hline \end{array} $$ Use the .01 significance level to determine whether there is a relationship between management level and environmental concern.
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