Question

Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. AIB Insurance randomly sampled 100 recently paid policies and determined the average age of clients in this sample to be 77.7 years with a standard deviation of 3.6. The 90% confidence interval for the true mean age of its life insurance policy holders is
A. (76.87, 80.33)
B. (72.5, 82.9)
C. (77.1, 78.3)
D. (74.1, 81.3)
E. (74.5, 80)

Answer 1

Answer:

$77.7-1.66\frac{3.6}{\sqrt{100}} =77.102$    

$77.7+1.66\frac{3.6}{\sqrt{100}} =78.30$    

And the best option would be:

C. (77.1, 78.3)

Step-by-step explanation:

Information given

$\bar X=77.7$ represent the sample mean

$\mu$ population mean (variable of interest)

s=3.6 represent the sample standard deviation

n=100 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 are given by:

$df=n-1=100-1=99$

Since the Confidence is 0.90 or 90%, the significance would be $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value for this case would be $t_{\alpha/2}=1.66$

And replacing we got:

$77.7-1.66\frac{3.6}{\sqrt{100}} =77.10$    

$77.7+1.66\frac{3.6}{\sqrt{100}} =78.30$    

And the best option would be:

C. (77.1, 78.3)