Determine the margin of error for aa 9595% confidence interval to estimate the p

Question

Determine the margin of error for

aa

9595%

confidence interval to estimate the population mean when s =

4242

for the sample sizes below.

a)

nequals=1212

b)

nequals=2828

c)

nequals=5050

a) The margin of error for

aa

9595%

confidence interval when

nequals=1212

is

nothing.

(Round to two decimal places as needed.)

b) The margin of error for

aa

9595%

confidence interval when

nequals=2828

is

nothing.

(Round to two decimal places as needed.)

c) The margin of error for

aa

9595%

confidence interval when

nequals=5050

is

nothing.

a)

nequals=1212

b)

nequals=2828

c)

nequals=5050

Explanation / Answer

Part a)

Here population standard deviations not given and the sample size is small so we use t-distribution to find confidence interval.

E = tc * ( s / sqrt (n))

tc is the critical value we find at 5% level of significance for 11 degrees of freedom. We get tc value as 2.201.

E = 2.201 * ( 42 / sqrt (12))

= 26.6857

Answer: 26.6857

Part b)

Here population standard deviations not given and the sample size is small so we use t-distribution to find confidence interval.

E = tc * ( s / sqrt (n))

tc is the critical value we find at 5% level of significance for 27 degrees of freedom. We get tc value as 2.052.

E = 2.052 * ( 42 / sqrt (28))

= 16.2872

Answer: 16.2872

Part c)

Here population standard deviations not given and the sample size is large so we use t-distribution to find confidence interval and we can use z-distribution also.

Using t-distribution:

E = tc * ( s / sqrt (n))

tc is the critical value we find at 5% level of significance for 49 degrees of freedom. We get tc value as 2.010.

E = 2.010 * ( 42 / sqrt (50))

= 11.9388

Answer: 11.9388

Using z-distribution:

E = zc * ( s / sqrt (n))

zc is the critical value we find at 5% level of significance is 1.96

E = 1.96 * ( 42 / sqrt (50))

= 11.6418

Answer: 11.6418

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