Question

Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of (i) 484 and (ii) 1600

(i) Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 484

(ii).

(ii) Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 1600

Answer #1

Solution :

Given that,

sample standard deviation = s = 90

1)

sample size = n = 484

Degrees of freedom = df = n - 1 = 484 - 1 = 483

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_{ /2,df} =
t_{0.025,483} = 1.965

Margin of error = E = t_{/2,df} * (s /n)

= 1.965* (90 / 484)

= 8.039

2)

sample size = n = 1600

Degrees of freedom = df = n - 1 = 1600 - 1 = 1599

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_{ /2,df} =
t_{0.025,1599} = 1.961

Margin of error = E = t_{/2,df} * (s /n)

= 1.961 * ( 90/ 1600)

= 4.412

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