Question

Assume that the weight of cereal in a "12.6 ounce box" is N(µ,0.42). The Food and Drug Association (FDA) allows only a small percentage of boxes to contain less than 12.6 ounces. You collect a random sample of n = 35 boxes and find that the sample mean is 12.8. Test the null hypothesis H0: µ = 13 against the alternative hypothesis HA: µ < 13 at level of significance α = 0.01

Answer 1

We reject the null hypothesis and accept the alternate hypothesis, that is, that the mean is less than 13 ounces.

Explanation:

The null hypothesis is:

`H_(0): \mu = 13`

The alternate hypotesis is:

`H_(a): \mu < 13`

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.

13 is tested at the null hypothesis:

This means that
`\mu = 13`

N(µ,0.42)

This means that the standard deviation is 0.42, that is,
`\sigma = 0.42`

You collect a random sample of n = 35 boxes and find that the sample mean is 12.8.

This means that
`n = 35, X = 12.8`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (12.8 - 13)/((0.42)/(√(35)))`

`z = -2.82`

Pvalue of the test:

The pvalue of the test is the pvalue of z = -2.82.

Looking at the z table, z = -2.82 has a pvalue of 0.0024

0.0024 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis, that is, that the mean is less than 13 ounces.