Question

Determine the upper limit of a 95% confidence level estimate of the mean when the population standard deviation (σ) is unknown and the sample size is 40, the sample mean is 110.27 and the sample standard deviation is 18.95. Determine the upper limit of a 95% confidence level estimate of the population proportion when the sample size is 200 customers, 35 of whom respond yes to a survey. In constructing a 99% confidence level estimate of the mean when the population standard deviation (σ) is known what will be your z score used in the formula?

Answer 1

a) At 95% confidence interval the critical value is t_{0.025, 39} = 2.023

The upper limit of the 95% confidence interval for population mean is

+ t_{0.025, 39} * s/

= 110.27 + 2.023 * 18.95/

= 110.27 + 6.06

= 116.33

b) = 35/200 = 0.175

At 95% confidence interval the critical value is z_0.025 = 1.96

The upper limit of the 95% confidence interval for population proportion is

+ z_0.025 * sqrt((1 - )/n)

= 0.175 + 1.96 * sqrt(0.175 * (1 - 0.175)/200)

= 0.175 + 0.053

= 0.228

c) For 99% confidence interval the critical value is z_0.005 = 2.58