A salesman who is on the road visiting clients thinks that on average he drives the same distance each day of the week. He keeps track of his mileage for several weeks and discovers that he averages 122 miles on Mondays, 203 miles on Tuesdays, 176 miles on Wednesdays, 181 miles on Thursdays, and 108 miles on Fridays. He wonders if this evidence contradicts his belief in a uniform distribution of miles across the days of the week. Explain why it is not appropriate to test his hypothesis using the chi-square goodness-of-fit test..Can someone help me solve this question...
To explain why it is not appropriate to use the chi-square goodness-of-fit test in this scenario, we need to understand the assumptions and requirements of this statistical test.
The chi-square goodness-of-fit test is used to determine if observed data differs significantly from the expected data, assuming a specified distribution. In this case, the salesman's hypothesis assumes a uniform distribution of mileage across the days of the week.
However, the chi-square goodness-of-fit test requires that the expected frequencies in each category be reasonably large (typically at least 5), and this assumption may not be met in this situation. The test works best when the sample size is large, preventing small expected frequencies that can lead to unreliable results.
In the given data, we have daily mileages that are quite diverse: ranging from 108 to 203. As a result, the expected frequencies for each day of the week would significantly differ from each other, making them potentially small and potentially violating the assumptions of the test. For example, if we equally distribute the total mileage across the days, the expected frequency might be around 158 for each day. Thus, the expected frequency for Monday (122) could be smaller than the recommended threshold of 5.
In this case, it would be more appropriate to use alternative statistical methods to analyze the data, such as comparing means or conducting a t-test to examine if the average distances differ significantly across the days of the week. These methods would allow a more appropriate and reliable analysis based on the given data.
To determine if the evidence contradicts the salesman's belief in a uniform distribution of miles across the days of the week, we typically use a chi-square goodness-of-fit test. However, in this case, the chi-square goodness-of-fit test is not appropriate for several reasons.
1. Sample Size: The chi-square goodness-of-fit test assumes that the sample size is large enough for the expected value of each category to be at least 5. In this case, the sample size is not mentioned, so it is unclear if this assumption is met.
2. Independence: The chi-square goodness-of-fit test assumes that the observations are independent of each other. However, in this scenario, the distances driven by the salesman on different days of the week are likely to be related because he may have similar clients or routes on certain days. Thus, the independence assumption is violated.
3. Categorical Data: The chi-square goodness-of-fit test is typically used for categorical data, where the observed frequencies are counted within mutually exclusive categories. In this scenario, the data is continuous (mileage) and not categorical.
Given these reasons, it is not appropriate to use the chi-square goodness-of-fit test in this case. Instead, you could consider other statistical tests such as a one-way ANOVA to analyze the differences in mileage across different days of the week.
This problem fits a multinomial experiment, not a goodness-of-fit test. The hypotheses would be something like the following:
Ho: There was no preference shown (equally distributed).
Ha: There was a preference shown (not equally distributed).
You can check out the assumptions of the two types of tests to determine the differences.