Question

A telco, using a sample of 85 accounts, found out that the average talk minutes per home user every month is 510 minutes. Lets assume that the standard deviation is 46 minutes. Create a 90%, 95% και 99% confidence interval for the average talk minutes.

Answer 1

Solution :

Given that,

(a)

At 90%

Z_/2 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (46 / 85)

= 8.21

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

- E < < + E

510 - 8.21 < < 510 + 8.21

501.79 < < 518.21

(501.79 , 518.21)

(b)

At 95%

Z_/2 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (46 / 85)

= 9.78

At 95% confidence interval estimate of the population mean is,

- E < < + E

510 - 9.78 < < 510 + 9.78

500.22 < < 519.78

(500.22 , 519.78)

(c)

At 99%

Z_/2 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (46 / 85)

= 12.85

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

- E < < + E

510 - 12.85 < < 510 + 12.85

497.15 < < 522.85

(497.15 , 522.85)