Question

Monica delivers pizza. Sometimes people tip and sometimes they don't. Below is a table of a random sample of 324 of Monica's deliveries.

Delivery Location Tip No tip Total
House 162 10 172
Apartment 80 9 89
Other
(trailer,
business, hotel) 56 7 63
Total 298 26 324



Is there a relationship between delivery location and whether or not Monica receives a tip?

1. The null hypothesis is
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2. What is the expected count for Apartment deliveries where the customer Tips?
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3. The expected count for House deliveries where the customer does not tip is 13.80. What is the component of the Chi-square statistic corresponding to this cell of the table?
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4. What are the degrees of freedom?
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5. The test statistic is 2.5. What do you conclude?
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Answer 1

1) The null hypothesis (H0) would be that there is no relationship between the delivery location and receiving a tip.

2) Expected Count is 81.85.

3) The component of the Chi-square statistic for the House deliveries is calculated using: (10 - 13.80)^2 / 13.80.

4) The degrees of freedom is 2.

5) If the test statistic is 2.5, without a given significance level, we cannot definitively conclude whether there is enough evidence to reject or not reject the null hypothesis.

Step-by-step explanation:

This question pertains to the assessment of whether there is a relationship between the delivery location and whether or not Monica receives a tip using the Chi-square test for independence.

1) The null hypothesis (H0) would be that there is no relationship between the delivery location and receiving a tip.

2) To calculate the expected count for Apartment deliveries where the customer tips, you would use the formula:

Expected Count = (Row Total * Column Total) / Grand Total.

So, it would be (89 * 298) / 324. = 81.85.

3) The component of the Chi-square statistic for the House deliveries where the customer does not tip is calculated using:

(Observed - Expected)^2 / Expected.

Here, it would be (10 - 13.80)^2 / 13.80.

4) The degrees of freedom (df) for a Chi-square test is calculated as (number of rows - 1) * (number of columns - 1).

In this case, df = (3-1) * (2-1)

= 2.

5) If the test statistic is 2.5, to reach a conclusion, we must compare this value against the critical value from the Chi-square distribution table for 2 degrees of freedom.

Without a given significance level, we cannot definitively conclude whether there is enough evidence to reject or not reject the null hypothesis.