Question

The average annual cost of Cable TV and Internet is unknown, with a standard deviation of $20. Suppose the average annual cost of Cable TV and Internet among participants in survey of a random sample of 64 subscribers is $99. Construct an 84% confidence interval to estimate the population average annual cost of Cable TV and Internet.

The lower limit is =

The upper limit is =

Answer 1

Solution :

Degrees of freedom = df = n - 1 = 63

At 84% confidence level the t is ,

= 1 - 84% = 1 - 0.84 = 0.16

/ 2 = 0.16 / 2 = 0.08

t /2,df = t0.08,63 = 1.422

Margin of error = E = t/2,df * (s /n)

= 1.422 * (20 / 64)

= 3.555

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

- E < < + E

99 - 3.555 < < 99 + 3.555

95.445 < < 102.555

The lower limit is = 95.445

The upper limit is = 102.555