Question

Yousure Inc. is a life insurance company. Recently, management conducted a survey of 100 customers who owned a particular type of policy. They found that the customers' mean age was 77.7 years, with a sample standard deviation of 36 years. Which of the following is the confidence interval for the true mean age of all customers who have this policy? 72.5 to 12 77.1 to 7.3 75.4 to 80 74.1 to 813 76.87 to 80.33

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 77.7

sample standard deviation = s = 3.6

sample size = n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ =1 - 0.90 = 0.10

$\alpha$/2 = 0.05

t$\alpha$/2,df = t_0.05,99 = 1.660

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.660 * ( 3.6 / $\sqrt$ 100)

Margin of error = E = 0.6

At 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

77.7 - 0.6 < $\mu$ < 77.7 + 0.6

77.1 < $\mu$ < 78.3

( 77.1 , 78.3 )

The 90% confidence interval for the true population mean : 77.1 to 78.3

Correct option : 77.1 to 78.3