Question

A restaurant owner believes diners at her restaurant are more likely to order a vegetarian entree while eating on a weekday than on the weekend. If this is true, she plans to offer a daily vegetarian special during the week. She collects a random sample of 88 diners eating in the restaurant on Monday, Tuesday, Wednesday, Thursday, or Friday and a random sample of 67 diners eating in the restaurant on Saturday or Sunday. She finds the 20% of weekday diners and 17% of weekend diners order a vegetarian option. Test her hypothesis.


(a) State the null and alternative hypotheses using correct notation.


(b) What type of test should you use? Explain how the conditions for this test are met.


(c) Choose an appropriate level of significance, and explain why you chose this value.


(d) Calculate the test statistic to two decimal places and the p-value to four decimal places.


(e) What is the result of the hypothesis test? Do you reject or fail to reject the claim?


(f) What are your conclusions. Should the restaurant owner offer the new vegetarian special?



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Answer 1

Solution:-

a)

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

Null hypothesis: P_Weekday< P_Weekend
Alternative hypothesis: P_Weekday > P_Weekend

Note that these hypotheses constitute a one-tailed test.

b) The test method is a two-proportion z-test.

c)

Formulate an analysis plan. For this analysis, the significance level is 0.05.

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 = 0.18703

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.06322
d)

z = (p₁ - p₂) / SE

z = 0.475

z = 0.48

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 one-tailed test, the P-value is the probability that the z-score is more than 0.475.

Thus, the P-value = 0.3156.

e) Do not reject the null hypothesis.

Interpret results. Since the P-value (0.3156) is greater than the significance level (0.05), we cannot reject the null hypothesis.

f)

From the above test we have sufficient evidence in the favor of the claim that diners in restaurant are more likely to order a vegetarian entree while eating on a weekday than on the weekend.