Problem 7 - Basic Statistics for Business and Economics with Student

Question

A group of department store buyers viewed a new line of dresses and gave their opinions of them. The results were: Because the largest number (47) indicated the new line is outstanding, the head designer thinks that this is a mandate to go into mass production of the dresses. The head sweeper (who somehow became involved in this) believes that there is not a clear mandate and claims that the opinions are evenly distributed among the six categories. He further states that the slight differences among the various counts are probably due to chance. Test the null hypothesis that there is no significant difference among the opinions of the buyers. Test at the .01 level of risk. Follow a formal approach; that is, state the null hypothesis, the alternate hypothesis, and so on.

Step-by-Step Solution

Answer

Test shows significant differences; potentially reject the null hypothesis.

Step 1: State the Hypotheses

The null hypothesis (\(H_0\)) is that the opinions are evenly distributed across the six categories. The alternative hypothesis (\(H_1\)) is that the opinions are not evenly distributed across the six categories.

Step 2: Collect and Categorize the Data

The data collected includes the number of buyers in each opinion category. List these numbers and note the total number of buyers. Assume the categories and corresponding counts are: Outstanding (47), Good (32), Average (21), Fair (15), Poor (10), Unsatisfactory (5).

Step 3: Calculate Expected Frequencies

If opinions are evenly distributed, each category should have the same number of buyers. Calculate the expected frequency by dividing the total number of responses by the number of categories. For example, suppose there are a total of 130 responses and 6 categories, then each category's expected frequency is \(\frac{130}{6} \approx 21.67\).

Step 4: Compute the Chi-Square Statistic

Use the formula \( X^2 = \sum \frac{(O_i - E_i)^2}{E_i} \) where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency for each category. Calculate this for all categories and sum the results to find the test statistic.

Step 5: Determine the Critical Value

The critical value for a \(\chi^2\) test with \(\alpha = 0.01\) and \(df = n - 1\) (where \(n\) is the number of categories) can be found using a Chi-Square distribution table. For 5 degrees of freedom, the critical value at the 0.01 level is approximately 15.09.

Step 6: Compare Chi-Square Statistic to Critical Value

Compare the calculated \(\chi^2\) statistic to the critical value from the table. If the statistic is greater than the critical value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Step 7: Conclusion

Based on the comparison, if the calculated \(\chi^2\) value exceeds the critical value, we reject the null hypothesis, suggesting a significant difference in opinions. Otherwise, we fail to reject the null hypothesis, indicating no significant difference.

Key Concepts

Chi-Square Test

The chi-square test is a statistical method used to determine if there is a significant difference between expected and observed frequencies in categorical data. This test helps us understand whether an observed result is due to chance or if it is statistically significant.
It is particularly useful when analyzing data in different categories to see if the proportion of observations in each category matches what we expect under the null hypothesis.

To perform a chi-square test, follow these basic steps:

• Calculate the expected frequencies for each category if the null hypothesis were true.
• Subtract the expected frequency from the observed frequency for each category, square the result, and divide by the expected frequency.
• Sum these values across all categories to obtain the chi-square statistic ( X^2 ).

This statistic is then compared to a critical value from the chi-square distribution table to decide on whether to reject the null hypothesis.

Null Hypothesis

The null hypothesis ( H_0 ) is a fundamental concept in hypothesis testing. It represents the assumption that there is no effect or no difference in the particular context of the study.
In the described exercise, the null hypothesis is that the opinions of the department store buyers are evenly distributed across the six opinion categories.

• This implies that any differences observed in the counts are simply due to random variation or chance.
• It's important to remember that the null hypothesis is assumed true until evidence suggests otherwise.

Testing this hypothesis allows us to statistically assess whether the observed distribution of opinions supports this idea or calls it into question. Only through rigorous testing and comparison to critical values can we decide to reject or fail to reject the null hypothesis.

Alternative Hypothesis

In contrast to the null hypothesis, the alternative hypothesis ( H_1 or H_a ) proposes that there is a significant effect or difference in the study. It suggests that the phenomenon observed is not simply due to chance.
In this particular exercise, the alternative hypothesis is that the opinions are not evenly distributed across the categories.

• If enough statistical evidence is found to support the alternative hypothesis, it usually leads us to "reject the null hypothesis."
• This means concluding that there is a significant difference in opinions among the categories in question.

Essentially, the alternative hypothesis is what the researcher aims to support and prove as true, indicating a meaningful signal in the data rather than random noise.

Significance Level

The significance level, often denoted by ext{alpha} ( ext{α} ), is a threshold used in hypothesis testing to determine if a result is statistically significant.
In our exercise, a significance level of 0.01 is used, meaning there's a 1% risk of rejecting the null hypothesis when it is actually true.

• This is known as a Type I error, where one wrongly concludes a difference or effect exists when it does not.
• Commonly used significance levels are 0.05, 0.01, and 0.001. Lower values indicate more stringent criteria for statistical significance.

The significance level is crucial because it helps in comparing the chi-square statistic against the critical value from the distribution table; if the statistic is larger, we reject the null hypothesis as there is strong evidence against it.