Question

A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligrams and a standard deviation of 1.4 milligrams. Is this in line with the manufacturer’s claim that the average nicotine content does not exceed 3.5 mg? Use a 0.01 level of significance and assume the distribution of nicotine contents to be normal.

Answer 1

0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.

Explanation:

Manufacturer’s claim that the average nicotine content does not exceed 3.5 mg

This means that the null hypothesis is given by:

`H_(0): \mu = 3.5`

And the alternate hypothesis is:

`H_(a): \mu > 3.5`

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.

3.5 is tested at the null hypothesis

This means that
`\mu = 3.5`

A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligrams and a standard deviation of 1.4 milligrams.

This means that
`n = 8, X = 4.2, \sigma = 1.4`

Value of the z-statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (4.2 - 3.5)/((1.4)/(√(8)))`

`z = 1.41`

Pvalue of the test:

We are testing if the mean is higher than 3.5.

The sample mean found is of 4.2, and we have to find the probability of finding a sample mean at least as large as this, which is 1 subtracted by the pvalue of z = 1.41.

z = 1.41 has a pvalue of 0.9207

1 - 0.9207 = 0.0793

0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.