Question

3. Fill in the blanks (show your work by drawing or using certain formula). a) (2 points) To form a 96% confidence interval for population proportion, 2.4 b) (2 points) To form a 98% confidence interval for population mean using a sample size of 26, to c) (2 points) For a 99% confidence interval with margin of error 0.03 with a prior estimate 0.65 of the sample proportion p in estimating population proportion p, we need to have sample size n= d) (2 points) For a 95% confidence interval with margin of error 0.05 and sample standard deviation s of 2.5 in estimating population mean i , we need to have sample size n=

Answer 1

solution:-

a) to form a 96% confidence interval for population proportion, zα/2 = 2.05

explanation:-

the z-value for 96% confidence level from standard normal distribution table zα/2 = 2.05

b) to form a 98% confidence interval for population mean using a sample size of 26 tα/2 = 2.485

explanation:-

here degree of freedom df = n - 1 = 26 - 1 = 25

we look into t-distribution table with df = 25 and with 98% confidence level

critical value tα/2 = 2.485

c) for a 99% confidence interval with margin of error 0.03 with a prior estimate 0.65 of the sample proportion in estimating population proporiton p we need to ahve sample size n = 1683

explanation:-

here given that margin of error E = 0.03 , P = 0.65

the z-value for 99% confidence is 2.58

sample size formula for population proportion

=> n = p*(1-p)*(z/E)^2

=> n = 0.65*(1-0.65)*(2.58/0.03)^2

=> n = 1683

d) for a 95% confidence interval with margin of error 0.05 and sample standard deviation s in estimating population x we need to have sample size n = 9604

explanation:-

here given that margin of error E = 0.05 , s = 2.5

the z-value for 95% confidence from standard normal table is 1.96

sample size formula for population mean

=> n = (z*s/E)^2

=> n = (1.96*2.5/0.05)^2

=> n = 9604