Question

(a) A senatorial candidate wants to estimate the proportion of people in his voting district who will favour a balanced-budget amendment, which is debatable. He wants this estimate to be within 0.03 of the population proportion for a 98%98% confidence interval. What is the sample size needed to produce this level of accuracy?

(b) A researcher wants to make a 95%95% confidence interval for the population mean. The population standard deviation is 25.5025.50. What sample size is required so that the sample mean will not differ from the population mean by more than 5.55.5 units?

Answer #1

Given data,

sample proportion is 50%

margin of error = 0.03

confidence level is 98%

sample size needed to produce this level of accuracy

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.02 is = 2.326

Sample Proportion = 0.5

ME = 0.03

n = ( 2.326 / 0.03 )^2 * 0.5*0.5

= 1502.8544 ~ 1503

b.

Given data,

standard deviation is 25.50

margin of error is 5.5

confidence level is 95%

sample size =n

Compute Sample Size

n = (Z a/2 * S.D / ME ) ^2

Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )

Standard Deviation ( S.D) = 25.5

ME =5.5

n = ( 1.96*25.5/5.5) ^2

= (49.98/5.5 ) ^2

= 82.579 ~ 83

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