Question

In: Statistics and Probability

A candy company fills a special package with individually wrapped pieces of candy. The number of...

A candy company fills a special package with individually wrapped pieces of candy. The number of pieces of candy per package varies because the package is sold by weight. The company wants to estimate the number of pieces per package. Inspectors randomly sample 100 packages of this candy and count the number of pieces in each package. They find that the sample mean number of pieces is 187. Assuming a population standard deviation of 8:

a. Construct a 98% confidence interval to estimate the mean number of pieces per package for the population.

b. In a new set of inspection, what is the sample size if inspectors want to be 95% confident of the results and want the error of the confidence interval (E) not to be more than 1.2?

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 187

Population standard deviation = = 8

Sample size = n = 100

a)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

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

= 2.326 * ( 8/ 100 )

= 1.86

At 98% confidence interval estimate of the population mean is,

- E < < + E

187 - 1.86 < < 187 + 1.86

185.14 < < 188.86

(185.14 , 188.86)

b)

Margin of error = E = 1.2

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

sample size = n = (Z/2* / E) 2

n = (1.96 * 8/ 1.2)2

n = 170.74

n = 171

Sample size = 171


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