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devlian [24]
3 years ago
7

Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. AIB Insuranc

e 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)
Mathematics
1 answer:
Radda [10]3 years ago
7 0

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)

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