Question

A simple random sample with n=56 provided a sample mean of 21.0 and a sample standard deviation of 4.7 .

a. Develop a 90% confidence interval for the population mean (to 1 decimal).

b. Develop a 95% confidence interval for the population mean (to 1 decimal).

c.Develop a 99% confidence interval for the population mean (to 1 decimal).

d. What happens to the margin of error and the confidence interval as the confidence level is increased?

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 21

sample standard deviation = s = 4.7

sample size = n = 56

Degrees of freedom = df = n - 1 = 55

a)

At 90% confidence level the t is ,

t_{$\alpha$ /2,df} = t_0.05,55 = 1.673

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

= 1.673 * (4.7 / $\sqrt$ 56)

= 1.05

The 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

21 - 1.05 < $\mu$ < 21 + 1.05

20.0 < $\mu$ < 22.1

( 20.0 , 22.1 )

b)

At 95% confidence level the t is ,

t_{$\alpha$ /2,df} = t_0.025,55 = 2.004

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

= 2.004 * (4.7 / $\sqrt$ 56)

= 1.2

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

21 - 1.2 < $\mu$ < 21 + 1.2

19.8 < $\mu$ < 22.2

( 19.8 , 22.8 )

c)

At 99% confidence level the t is ,

t_{$\alpha$ /2,df} = t_0.005,55 = 2.668

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

= 2.668 * (4.7 / $\sqrt$ 56)

= 1.7

The 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

21 - 1.7 < $\mu$ < 21 + 1.7

19.3 < $\mu$ < 22.7

( 19.3 , 22.7 )

d)

The level of confidence increases , then the margin of error is increased and the confidence interval is wider .