Question

A candy company fills a 20-ounce package of Halloween candy 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 120 packages of this candy and count the number of pieces in each package. They find that the sample mean number of pieces is 18.72 and the sample standard deviation is 0.8735. Construct a 99% confidence interval to estimate the mean number of pieces per package for the population of 20-ounce Halloween candy packages. Interpret your interval.

Answer 1

Solution:

The 99% confidence interval for population mean is given as follows:

Where, x̄ is sample mean, s is sample standard deviation, n is sample size and t_{(0.01/2, n - 1)} is critical t value to construct 99% confidence interval.

We have, x̄ = 18.72, s = 0.8735 and n = 120

Using t-table we get, t_{(0.01/2, 120 - 1)} = 2.6178

Hence, 99% confidence interval to estimate the mean number of pieces per package for the population of 20-ounce Halloween candy packages is,

The 99% confidence interval to estimate the mean number of pieces per package for the population of 20-ounce Halloween candy packages is (18.5113, 18.9287).

Interpretation : We are 99% confident that true value of the mean number number of pieces per package for the population of 20-ounce Halloween candy packages lies within the confidence limits of the 99% confidence interval.

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