Question

The owner of a specialty coffee shop wants to study coffee purchasing habits of customers at her shop. She selects
a random sample of 60 customers during a certain week. Data are available in the worksheet labeled “Problem 2”
in the spreadsheet.

1. Construct a 90% confidence interval on the proportion of customers who say they “definitely will”
recommend the specialty coffee shop to family and friends.

Problem 2 Data Set

Note: Y is an indicator variable, i.e., if Y=1, then customer said they "definitely would" recommend specialty shop to family and friends, and Y=0 otherwise.

Customer	Y
1	0
2	1
3	0
4	0
5	0
6	0
7	0
8	1
9	1
10	1
11	0
12	1
13	0
14	1
15	0
16	1
17	1
18	1
19	0
20	1
21	1
22	0
23	0
24	0
25	0
26	1
27	0
28	0
29	1
30	0
31	1
32	1
33	0
34	0
35	0
36	1
37	0
38	1
39	1
40	1
41	1
42	0
43	1
44	0
45	1
46	0
47	1
48	0
49	0
50	0
51	1
52	1
53	1
54	0
55	1
56	0
57	0
58	0
59	1
60	1

Answer 1

Customer	Y
1	0
2	1
3	0
4	0
5	0
6	0
7	0
8	1
9	1
10	1
11	0
12	1
13	0
14	1
15	0
16	1
17	1
18	1
19	0
20	1
21	1
22	0
23	0
24	0
25	0
26	1
27	0
28	0
29	1
30	0
31	1
32	1
33	0
34	0
35	0
36	1
37	0
38	1
39	1
40	1
41	1
42	0
43	1
44	0
45	1
46	0
47	1
48	0
49	0
50	0
51	1
52	1
53	1
54	0
55	1
56	0
57	0
58	0
59	1
60	1
Total count	60
Yes Count	29

proportion

0.483333

We need to construct the 90% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X =	29
Sample Size N =	60

The sample proportion is computed as follows, based on the sample size N=60 and the number of favorable cases X = 29:

The critical value for α=0.1 is . The corresponding confidence interval is computed as shown below:

CI(Proportion)​=(0.377,0.589)​

Therefore, based on the data provided, the 90% confidence interval for the population proportion is 0.377<p<0.589, which indicates that we are 90% confident that the true population proportion p is contained by the interval (0.377, 0.589).

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