A random sample of the amounts for 22purchases was taken. The mean was $49.8649.

Question

A random sample of the amounts for 22purchases was taken. The mean was $49.8649.86, the standard deviation was $23.9323.93, and the margin of error for a 95% confidence interval was $10.6110.61. Assume that t *n1equals=2.0 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?

a) The new sample size should be

(Round up to the nearest integer.)

b) The new sample size should be

Explanation / Answer

a. Margin of error = E=(z*sd)/sqrt(n) n=(z*sd/E)^2 for E=5, z=1.96 for 95% CI and sd=23.93 Hence n=((1.96*23.93)/5)^2=88 b. Here E=1 Hence n=((1.96*23.93)/1)^2=2200

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