2a. Describe the sample and population you are studying.
I am studying customers' satisfaction level with their car insurance plan.
2b. Calculate the sample mean and sample standard deviation for
your variable. Keep at least four decimal places.
2c. Calculate the standard error and the margin of error for a 90%
confidence interval. Keep at least four decimal places.
Data:
36 |
18 |
66 |
43 |
28 |
39 |
47 |
40 |
24 |
46 |
48 |
57 |
36 |
58 |
39 |
62 |
43 |
65 |
74 |
36 |
39 |
44 |
61 |
50 |
47 |
63 |
60 |
38 |
45 |
51 |
55 |
46 |
68 |
32 |
42 |
38 |
61 |
45 |
31 |
32 |
44 |
30 |
29 |
62 |
49 |
54 |
64 |
38 |
49 |
55 |
28 |
53 |
55 |
52 |
50 |
54 |
76 |
28 |
49 |
70 |
29 |
34 |
77 |
40 |
50 |
40 |
56 |
54 |
36 |
51 |
42 |
71 |
45 |
53 |
55 |
37 |
51 |
36 |
39 |
36 |
51 |
40 |
51 |
52 |
53 |
33 |
66 |
37 |
76 |
67 |
55 |
46 |
a)
sample is random people selected for the study
population is the total customers who has availed the car insurance
b)
sample mean, xbar = 47.837
sample standard deviation, s = 12.8553
c)
sample size, n = 92
SE = 12.8553/sqrt(92) = 1.3403
degrees of freedom, df = n - 1 = 91
Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.662
ME = tc * s/sqrt(n)
ME = 1.662 * 12.8553/sqrt(92)
ME = 2.2275
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (47.837 - 1.662 * 12.8553/sqrt(92) , 47.837 + 1.662 *
12.8553/sqrt(92))
CI = (45.6095 , 50.0645)
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