Question

A large hotel chain claims that 96% of its customers leave satisfied with their stay. The results of a customer service study are contained below (this sample is also contained in the DATA 1 tab of the downloaded excel file). . At a 1% significance level, is there evidence that the proportion of satisfied customers is actually less than 96%?

Satisfied with hotel stay?
YES	YES	YES	YES	YES	YES	YES
YES	YES	NO	YES	YES	YES	YES
YES	YES	YES	NO	YES	YES	YES
NO	YES	YES	YES	YES	YES	YES
YES	YES	YES	YES	NO	YES	YES
NO	YES	YES	YES	YES	YES	NO
YES	YES	YES	YES	YES	YES	YES
YES	YES	YES	YES	YES	YES	YES
YES	NO	YES	YES	YES	YES	YES
YES	YES	YES	YES	YES	YES	YES
YES	YES	YES	YES	NO	YES	YES
YES	YES	YES	YES	NO	YES	YES
YES	YES	YES	YES	YES	YES	YES
YES	YES	YES	YES	YES	YES	YES
YES	YES	YES	YES	YES	YES	YES
YES	NO	YES	YES	YES	YES	YES

a) Properly state the hypotheses:(1 mark)

b) Calculate the p-value (SHOW ALL WORK):

IN ORDER TO SHOW YOUR WORK FOR PART B...CREATE A TABLE THAT MIMICS THE EXCEL FUNCTION YOU ARE USING...YOUR ANSWER SHOULD INCLUDE THE FOLLOWING:

1) FUNCTION NAME:

2) FIELD 1 ENTRY:

3) FIELD 2 ENTRY:

4) FIELD 3 ENTRY:

5) FIELD 4 ENTRY:

6) YOUR FINAL ANSWER:

Answer 1

No. of Yes: x = 102

Total: n = 112

p̂ = x/n = 102/112 = 0.91, p = 0.96

Level of significance = 0.01

a) H₀: p > 0.96, Proportion of satisfied customers is not less than 96%

H₁: p < 0.96, Proportion of satisfied customers is actually less than 96%

b) Test statistic = (p̂-p)/(p*(1-p)/n)^0.5 = (0.91-0.96)/(0.96*(1-0.96)/112)^0.5 = -2.7

p-value (Using Excel function NORM.S.DIST(test statistic,cumulative)) = NORM.S.DIST(-2.7,TRUE) = 0.003

Since p-value is less than 0.01, we reject the null hypothesis.

So, p < 0.96, proportion of satisfied customers is actually less than 96%.