Question

A machine that is programmed to package 1.90 pounds of cereal is being tested for its accuracy in a sample of 53 cereal boxes, the sample mean filling weight is calculated as 1.92 pounds. The population standard deviation is known to be 0.06 pounds. Find the 95% confidence interval for the population mean filling weight.

[1.9038, 1.9362]

[1.8219, 1.9567]

[1.8646, 1.9800]

[1.8449, 1.9969]

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1.92

Population standard deviation =    = 0.06

Sample size = n = 53

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.05 = 1.96}

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

= 1.96 * ( 0.06 /  53 )

= 0.0162

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

- E < < + E

1.92 - 0.0162  <   < 1.92 + 0.0162

1.9038 <   < 1.9362

( 1.9038 , 1.9362 )

At 95% confidence interval estimate of the population mean is : ( 1.9038 , 1.9362 )

Correct option : [ 1.9038 , 1.9362 ]