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Elza [17]
3 years ago
15

A random sample of the amounts for 22 purchases was taken. The mean was ​$42.97​, the standard deviation was ​$22.82​, and the m

argin 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​?
Mathematics
1 answer:
Illusion [34]3 years ago
8 0

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

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