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

Answer:

$19.6-2.42\frac{5.8}{\sqrt{42}}=17.43$    

$19.6+2.42\frac{5.8}{\sqrt{42}}=21.77$    

And the best option for this case would be:

a. (17.5, 21.7)

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$19.6-2.42\frac{5.8}{\sqrt{42}}=17.43$    

$19.6+2.42\frac{5.8}{\sqrt{42}}=21.77$    

And the best option for this case would be:

a. (17.5, 21.7)