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. A random sample of 120 packages of this candy was inspected and number of pieces in each package was counted. Mean number of pieces in the sample was found to be 18.72. Assuming a population standard deviation of 0.8735, construct 99% confidence interval to estimate the mean number of pieces per package for the population.

Answer 1

Answer:

(18.51467,18.92532)

Step-by-step explanation:

we know that sample size, n, is 120

sample mean is 18.72

population standard deviation is 0.8735

confidence level is 99%

For a population with unknown mean $\beta$  and known standard deviation,S.D, known sample mean S.M, a confidence interval for the population mean, based on a simple random sample (SRS) of size n, is

$(S.M+Z*\frac{S.D}{\sqrt{n}},  S.M+Z*\frac{S.D}{\sqrt{n} })$

where Z  is the upper (1-C)/2 critical value for the standard normal distribution

z=2.575 for the 99% confidence level

therefore the confidence interval is;

$(18.72-2.575*\frac{0.8735}{\sqrt{120} }, 18.72-2.575*\frac{0.8735}{\sqrt{120} })\\\\(18.72-0.2053286,18.72+0.2053286)\\\\(18.51467,18.92532)$

This means that there is a 99% certainty that mean number of pieces per package for the population lies within the interval (18.51467,18.92532)