34.2k views
0 votes
The average weight of a package of rolled oats is supposed to be at most 18 ounces. A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces. (a) at the 5 percent level of significance, is the true mean larger than the specification

User Shivid
by
7.7k points

1 Answer

4 votes

Answer:

Pvalue of 0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

Explanation:

The average weight of a package of rolled oats is supposed to be at most 18 ounces.

This means that the null hypothesis is:


H_(0): \mu \leq 18

Is the true mean larger than the specification?

Due to the question asked, the alternate hypothesis is:


H_(a): \mu > 18

The test statistic is:


z = (X - \mu)/((\sigma)/(√(n)))

In which X is the sample mean,
\mu is the value tested at the null hypothesis,
\sigma is the standard deviation and n is the size of the sample.

18 is tested at the null hypothesis:

This means that
\mu = 18

A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces.

This means that
n = 18, X = 18.2, \sigma = 0.5

Value of the test statistic:


z = (X - \mu)/((\sigma)/(√(n)))


z = (18.2 - 18)/((0.5)/(√(18)))


z = 1.7

Pvalue of the test:

Probability of finding a sample mean above 18.2, which is 1 subtracted by the pvalue of z = 1.7.

Looking at the z-table, z = 1.7 has a pvalue of 0.9554.

1 - 0.9554 = 0.0446

0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

User Jkebinger
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories