Question

Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 92, with a sample size of (a) 400,(b) 1800. What is the effect of the sample size? 2. The margin of error for a 95% confidence interval with a sample size of 400 is (Round to the nearest tenth as needed.) b. The margin of error for a 90% confidence interval with a sample size of 1600 is (Round to the nearest tenth as needed) What is the effect of the sample size? O A As the sample size increases, the margin of error becomes smaller. OB. As the sample size increases, the margin of error does not change. O c. As the sample size increases, the margin of error becomes larger

Answer 1

solution:

a) Given that

Sample size(n) = 400

Sample standard deviation (s) = 92

we know that

Margin of error = Critical value * Standard error = $Z*(\frac{s}{\sqrt n})$

For 95% confidence level , $\alpha$ = 1 - CL = 1 - 0.95 = 0.05

Critical value : Z($\alpha$/2) = Z(0.025) = 1.96

$\therefore$ Margin of error = $Z*(\frac{s}{\sqrt n})$ = $1.96 * \frac{92}{\sqrt 400}$ = 9.016 =~ 9.0

b) Given that

Sample size(n) = 1600

Sample standard deviation (s) = 92

we know that

Margin of error = Critical value * Standard error = $Z*(\frac{s}{\sqrt n})$

For 95% confidence level , $\alpha$ = 1 - CL = 1 - 0.95 = 0.05

Critical value : Z($\alpha$/2) = Z(0.025) = 1.96

$\therefore$ Margin of error = $Z*(\frac{s}{\sqrt n})$ = $1.96 * \frac{92}{\sqrt 1600}$ = 4.508 =~ 4.5

c) we can observe that

when the sample size increases , the margin of error decreases

So,Option-A is correct.