Answer:
[tex]19.6-2.42\frac{5.8}{\sqrt{42}}=17.43[/tex]
[tex]19.6+2.42\frac{5.8}{\sqrt{42}}=21.77[/tex]
And the best option for this case would be:
a. (17.5, 21.7)
Step-by-step explanation:
Information given
[tex]\bar X= 19.6[/tex] represent the sample mean
[tex]\mu[/tex] population mean
[tex]\sigma= 5.8[/tex] represent the population deviation
n=42 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom, given by:
[tex]df=n-1=42-1=41[/tex]
Since the Confidence is 0.98 or 98%, the significance would be [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.1[/tex], and the critical value would be [tex]t_{\alpha/2}=2.42[/tex]
Replacing we got:
[tex]19.6-2.42\frac{5.8}{\sqrt{42}}=17.43[/tex]
[tex]19.6+2.42\frac{5.8}{\sqrt{42}}=21.77[/tex]
And the best option for this case would be:
a. (17.5, 21.7)
Answer:
The 98% confidence interval for the population mean is between 17.5 hours and 21.7 hours.
