Question

You measure 43 textbooks' weights, and find they have a mean weight of 56 ounces. Assume the population standard deviation is 11.3 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places.

< μμ <

Answer 1

Solution :

Given that,

= 56

= 11.3

n = 43

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * ( 11.3/ 43)

= 2.83

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

- E < < + E

56 - 2.83 < < 56 + 2.83

53.17 < < 58.83

90% confidence interval for the true population mean textbook weight.

53.17 < < 58.83

(53.17 , 58.83)