Question

A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$45.95 and the standard deviation is ​$22.34.

20.22. 20.31. 20.62

2.64. 61.94. 61.97

42.96. 43.19. 44.21

24.35. 44.71. 65.12

65.57. 49.36. 49.75

86.31. 33.45. 34.70

88.05. 59.60

​a) Construct a 90% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met.

​b) How large is the margin of​ error?

​c) How would the confidence interval change if you had assumed that the standard deviation was known to be ​$23?

Answer 1

Solution

a)

Summarizing the information we get

n = 20

t(0.05, 19) = 1.729

=> 45.9515+- 1.729133*22.3439/sqrt(20)

= 37.3123, 54.5907

b)

= 1.729133*22.3439/sqrt(20)

= 8.6392

c)

At 90% zcrit is 1.645

= 45.9515+-1.645*23/sqrt(30)

= 39.0438, 52.8592