The owner of a restaurant serving Continental-style entrées has the business obj

Question

The owner of a restaurant serving Continental-style entrées has the business objective of learning more about the patterns of patrons demand during the Friday-to-Sunday weekend time period. She decided to study the demand for dessert during this time period. In addition to studying whether a dessert was ordered, she will study the gender of the individual and whether a beef entree was ordered. Data was collected form 630 customers and organzied in the following contingency tables:

GENDER
DESSERT Male   Female    Total
Yes 50 96 146
No 250 234 484
Total 300 330 630


BEEF ENTRÉE
DESSERT Yes       No         Total
Yes 74 68 142
No 123 365 488
Total 197 433 630

a. At the 0.05 level of significance, is there evidence of a difference between males and females in the proporation who ordered dessert?

b. At the 0.05 level of significance, is there evidence of a difference in the proportion who ordered dessert based on whether a beef entree has been ordered?

Explanation / Answer

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.23175
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.03366

z = (p1 - p2) / SE

z = - 0.54

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.54 or greater than 0.54.

Thus, the P-value = 0.5892

Interpret results. Since the P-value (0.5892) is greater than the significance level (0.05), we have to accept the null hypothesis.

At the 0.05 level of significance, there not evidence of a difference between males and females in the proporation who ordered dessert.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.2254
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.03591

z = (p1 - p2) / SE

z = 6.09

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 6.09 or greater than 6.09.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

At the 0.05 level of significance, there is evidence of a difference in the proportion who ordered dessert based on whether a beef entree has been ordered.

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