Question

Suppose a random sample of size 17 was taken from a normally distributed population, and the sample standard deviation was calculated to be s = 5.0.

a) Calculate the margin of error for a 95% confidence interval for the population mean.

Round your response to at least 3 decimal places.

   

b) Calculate the margin of error for a 90% confidence interval for the population mean.

Round your response to at least 3 decimal places.

Answer 1

Solution :

Given that,

sample standard deviation = s = 5.0

sample size = n = 17

Degrees of freedom = df = n - 1 = 17-1= 16

a) At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t _{$\alpha$ /2,df} = t_0.025,16 = 2.120

Margin of error = E = t _{$\alpha$ /2,df} * (s / $\sqrt$ n)

= 2.120 * (5.0 / $\sqrt$ 17)

E = 2.571

margin of error for a 95% confidence interval for the population mean is 2.571

b) At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.9 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

t _{$\alpha$ /2,df} = t_0.05,16 = 1.746

Margin of error = E = t _{$\alpha$ /2,df} * (s / $\sqrt$ n)

= 1.746 * (5.0 / $\sqrt$ 17)

E = 2.117

margin of error for a 90% confidence interval for the population mean is 2.117