Question

In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6. Assume the population standard deviation is 5.8 hours. Find the 98% confidence interval for the population mean.

a. (17.5, 21.7)
b. (14.1, 23.2)
c. (18.3, 20.9)
d. (19.1, 20.4)

Answer 1

The 98% confidence interval for the population mean is between 17.5 hours and 21.7 hours.

Answer 2

`19.6-2.42(5.8)/(√(42))=17.43`

`19.6+2.42(5.8)/(√(42))=21.77`

And the best option for this case would be:

a. (17.5, 21.7)

Explanation:

Information given

`\bar X= 19.6` represent the sample mean

`\mu` population mean

`\sigma= 5.8` represent the population deviation

n=42 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom, given by:

`df=n-1=42-1=41`

Since the Confidence is 0.98 or 98%, the significance would be
`\alpha=0.02` and
`\alpha/2 =0.1`, and the critical value would be
`t_(\alpha/2)=2.42`

Replacing we got:

`19.6-2.42(5.8)/(√(42))=17.43`

`19.6+2.42(5.8)/(√(42))=21.77`

And the best option for this case would be:

a. (17.5, 21.7)