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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

Sagot :

Answer:

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

Step-by-step explanation:

The null hypothesis is:

[tex]H_{0}: \mu = 13[/tex]

The alternate hypotesis is:

[tex]H_{a}: \mu < 13[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

13 is tested at the null hypothesis:

This means that [tex]\mu = 13[/tex]

N(µ,0.42)

This means that the standard deviation is 0.42, that is, [tex]\sigma = 0.42[/tex]

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

This means that [tex]n = 35, X = 12.8[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{12.8 - 13}{\frac{0.42}{\sqrt{35}}}[/tex]

[tex]z = -2.82[/tex]

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.