Question

The sugar content in a one-cup serving of a certain breakfast cereal was measured for a sample of 140 servings. The average was 11.9 g and the standard deviation was 1.1 g.

Find a 95% confidence interval for the mean sugar content. Round the answers to three decimal places.

The 95% confidence interval is ( ), ( )

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 11.9

Population standard deviation =    = 1.1

Sample size n =140

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96* (1.1 / 140 )

= 0.182
At 95% confidence interval estimate of the population mean
is,

- E < < + E

11.9- 0.182 <   < 11.9 + 0.182

11.718<   < 12.082

( 11.718, 12.082 )