Question

A random sample of the amounts for 22 purchases was taken. The mean was ​$42.97​, the standard deviation was ​$22.82​, and the margin of error for a 95​% confidence interval was ​$10.12. Assume that t Subscript n minus 1 Superscript starequals2.0 for the​ 95% confidence intervals. ​a) To reduce the margin of error to about ​$5​, how large would the sample size have to​ be? ​b) How large would the sample size have to be to reduce the margin of error to ​$1.0​?

Answer 1

Answer: a) 84 and b ) 2084

Explanation:

Given : Sample standard deviation : s= $22.82

(Population standard deviation is unknown ) , so we use t-test.

Critical value or the​ 95% confidence intervals :
`t_(n-1)*=2.0`

Formula to find the sample size :

`n=((t^*\cdot s)/(E))^2`

a) E = 5

`n=(((2)\cdot 22.82)/(5))^2`

`n=(9.128)^2=83.320384\approx84`

i.e. Required sample size : n= 84

b) E = 1

`n=(((2)\cdot 22.82)/(1))^2`

`n=(45.64)^2=2083.0096\approx2084`

i.e. Required sample size : n= 2084