Posts Tagged ‘confidence interval’

Construct the confidence interval for the population mean

 

Construct the confidence interval for the population mean .

c = 0.90, \overline{x}s = 17.8, s = 12.0, and n = 110

Solution

Use the Formula E =  z_{c}\frac{\sigma }{\sqrt{n}}

to find the margin of error, where z_{c} is the z-score corresponding to an area of c,

\sigma is the population standard deviation and

n is the sample size.

When n\geq 30 you can use s in place of \sigma, where s is the sample standard deviation.

z_{c}  = \frac{1}{2}(1-c).

Substitute the value of c and evaluate to find the area in each tail.

 

z_{c}  =  \frac{1}{2}(1-c)

=\frac{1}{2}(1-0.90) = 0.05

Use a standard normal table to find the positive critical value the corresponds to a tail area of 0.05

z_{c} = 1.645

The population standard deviation \sigma is unknown but because n\geq30, use s in place of \sigma.

Substitute the value z_{c} = 1.645,  \sigma  \approx s \approx12.0, and n = 110 to find E.

 

E = z_{c}\frac{\sigma }{\sqrt{n}}

1.645\times \frac{12}{\sqrt{110}}

\approx1.9

Use the margin of error to find the left endpoint.

Left endpoint = \overline{x} – E

= 17.8 – 1.9 = 15.9

Now find the right endpoint

Right endpoint = \overline{x} + E

= 17.8 + 1.9 = 19.7

Therefore,  a  90% confidence interval for \mu is (15.9, 19.7).