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

Step-by-step 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 :

$n=(\dfrac{t^*\cdot s}{E})^2$

a) E = 5

$n=(\dfrac{(2)\cdot 22.82}{5})^2$

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

i.e. Required sample size : n= 84

b)  E = 1

$n=(\dfrac{(2)\cdot 22.82}{1})^2$

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

i.e. Required sample size : n= 2084