Question

1. As the new manager of a small convenience store, you want to understand the shopping patterns of your customers. You randomly sample 20 purchases from yesterday’s records (all purchases in U.S. dollars):

39.05 2.73 32.92 47.51 37.91 34.35 64.48 51.96 56.95 81.58 47.80 11.72 21.57 40.83 38.24 32.98 75.16 74.30 47.54 65.62

1. Find the sample mean and the standard deviation.
2. What is the standard error of the mean?
3. Construct a 90% confidence interval for the mean age of all customers, assuming that the distribution of age is normal.
4. How large is the margin of error

Answer 1

Sample size = n Part-A: Sample mean 20 905.20 = 45.26 Sample Standard Deviation: 8117.213 20120.669 Part-B: Standard error of the mean: s 20.669 = 4.622 Part-C: Degrees of freedom =df = n-1 20-1 19 For 90% confidence interval, α 1-0.90 0.10 a/2-0.10/): 0.05 From t-distribution table, the value of t having an area of (a/2 = 0.05) in its upper tail and for (df = 19) is given by = to.os - 1.729 Margin of Error 0E =10.05 → MoE = 1.729 × → MoE = 7 .992 20.669 V20