Question

The mean and standard deviation of the sample of 50 customer satisfaction ratings are 60 and 2.65. Set up a 99 percent confidence interval for μ (the true mean of the customer satisfaction ratings). Interpret this interval.

Answer 1

Solution :

degrees of freedom = n - 1 = 50 - 1 = 49

t$\alpha$/2,df = t_0.005,49 = 2.680

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

= 2.680 * ( 2.65 / $\sqrt$ 50)

Margin of error = E = 1.00

The 99% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 60  ± 1.00

= ( 59.00, 61.00 )

We are 99% confident that the true mean of  the customer satisfaction ratings between 59.00 and 61.00.