Question

The sample size required to estimate, at 95% confidence, a population mean with a maximum allowable margin of error of +/- 1.5 when the population standard deviation is 5 is ____ observations.

The sample size required to estimate, at 90% confidence, a population proportion with a maximum allowable margin of error of +/- 3.00 percentage points is ____ observations.

Answer #1

Solution :

Given that,

standard deviation = = 5

margin of error = E = 1.5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_{/2}
= Z_{0.025} = 1.96

Sample size = n = ((Z_{/2}
*
) / E)^{2}

= ((1.96 * 5) / 1.5)^{2}

= 42.68 = 43

Sample size = 43

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 3% = 0.03

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_{/2}
= Z_{0.05} = 1.645

sample size = n = (Z_{ / 2} / E )^{2} *
* (1 - )

= (1.645 / 0.03)^{2} * 0.5 * 0.5

= 751.67 = 752

sample size = 752

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