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 Final_SU2020_Data_Sets.xlsx. 1. At the ? = 0.1 level of significance, is there evidence that more than 50% of the customers say they “definitely will” recommend the specialty coffee shop to family and friends? 2. Construct a 90% confidence interval on the proportion of customers who say they “definitely will” recommend the specialty coffee shop to family and friends.

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

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.

Answer 1

Solution

Final answers are given below. Back-up Theory and Details of calculations follow at the end.

From the given data, 29 customers out of 60 sampled say they “definitely will” recommend, i.e., yes or 1 in the given table.

Part (1)

At the ? = 0.1 level of significance, there is no evidence that more than 50% of the customers say they “definitely will” recommend the specialty coffee shop to family and friends. Answer 1

Part (2)

90% confidence interval on the proportion of customers who say they “definitely will” recommend the specialty coffee shop to family and friends is: [0.38, 0.59] Answer 2

Back-up Theory and Details of calculations

Part (1)

Let X = number of customers say they “definitely will” recommend the specialty coffee shop to family and friends.

Then, X ~ B(n, p),

where

n = sample size

and

p = probability that customers say they “definitely will” recommend the specialty coffee shop to family and friends, which is also equal to the population proportion.

Claim :

More than 50% of the customers say they “definitely will” recommend the specialty coffee shop to family and friends.

Hypotheses:

Null H₀ : p = p₀ = 0.5 Vs Alternative H_A : p > 0.5 [claim]

Test Procedure

Z = (p_nat - p₀)/√{p₀(1 - p₀)/n}

where

p_nat = sample proportion and

n = sample size.

Calculations:

p0	0.5
n	60
x	29
phat	0.4833
Zcal	-0.2582
α	0.1
Zcrit	1.2816
p-value	0.6019

Distribution, Significance Level, α, Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided

np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α%, Critical Value = upper α% of N(0, 1), and

p-value = P(Z > Zcal)

Using Excel Functions: Statistical NORMSINV and NORMDIST, critical value and p-value are found to be as shown in the above table.

Decision:

Since Zcal < Zcrit, or equivalently since p-value > α, H₀ is accepted.

Conclusion :

There is not enough evidence to suggest that the claim is valid.

Part (2)

100(1 - α) % Confidence Interval for the population proportion, p is:

p_hat ± MoE, ………………………………………………………..…………. (1)

where

MoE = Z_α/2 x SE(p_hat) …………………………………………………….. (1a)

and

SE(p_hat) = √{p_hat (1 – p_hat)/n}…. ………………………..….……………..(1b)

with

Z_α/2 is the upper (α/2)% point of N(0, 1),

p_hat = sample proportion, and

n = sample size.

Calculations

n	60
X	29
p' = phat	0.4833
F = p'(1-p')/n	0.0042
sqrtF	0.0645
α	0.1
1 - (α/2)	0.95
Zα/2	1.6449
MoE	0.1061
LB	0.3772
UB	0.5894

DONE