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

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