Question

Determine the margin of error for

a

95%

confidence interval to estimate the population mean when s​ =

37

for the sample sizes below.

​a)	nequals=15
​b)	nequals=34
​c)	nequals=46

​a) The margin of error for

a

95​%

confidence interval when

nequals=15

is

nothing.

​(Round to two decimal places as​ needed.)

​b) The margin of error for

a

95%

confidence interval when

nequals=34

is

nothing.

​(Round to two decimal places as​ needed.)

​c) The margin of error for

a

95​%

confidence interval when

nequals=46

is

nothing.

​(Round to two decimal places as​ needed.)

Answer 1

a) n = 15

df = 15 - 1 = 14

T score for 95% confidence interval = t_0.025,14 = 2.145

Margin of error = t_0.025,14 * s / sqrt(n) = 2.145 * 37 / sqrt(15) = 20.49

b) n = 34

df = 34 - 1 = 33

T score for 95% confidence interval = t_0.025,33 = 2.035

Margin of error = t_0.025,33 * s / sqrt(n) = 2.035 * 37 / sqrt(34) = 12.91

c) n = 46

df = 46 - 1 = 45

T score for 95% confidence interval = t_0.025,45 = 2.014

Margin of error = t_0.025,45 * s / sqrt(n) = 2.014 * 37 / sqrt(46) = 10.99