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:

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

Step-by-step explanation:

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

This means that the null hypothesis is given by:

[tex]H_{0}: \mu = 3.5[/tex]

And the alternate hypothesis is:

[tex]H_{a}: \mu > 3.5[/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.

3.5 is tested at the null hypothesis

This means that [tex]\mu = 3.5[/tex]

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 [tex]n = 8, X = 4.2, \sigma = 1.4[/tex]

Value of the z-statistic:

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

[tex]z = \frac{4.2 - 3.5}{\frac{1.4}{\sqrt{8}}}[/tex]

[tex]z = 1.41[/tex]

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.

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