Question

A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligram and a standard deviation of 1.4 milligrams. Is this in line with manufacturer’s claim that the average nicotine content does exceed 3.5 milligram? Use a 0.1 level of significance and assume the distribution of nicotine contents to be normal. The value of the test statistic for testing this hypothesis is _____

Answer 1

To test the manufacturer's claim that the average nicotine content exceeds 3.5 milligrams, we can set up our hypothesis tests as follows:

- Null hypothesis (\(H_0\)): \(\mu \leq 3.5\) (The average nicotine content is less than or equal to 3.5 mg.)
- Alternative hypothesis (\(H_a\)): \(\mu > 3.5\) (The average nicotine content is greater than 3.5 mg.)

Given:
- Sample size (\(n\)) = 8
- Sample mean (\(\bar{x}\)) = 4.2 mg
- Sample standard deviation (\(s\)) = 1.4 mg
- Significance level (\(\alpha\)) = 0.1

We use the t-test for the sample mean since the sample size is small and we are assuming the distribution is normal.

The test statistic for the one-sample t-test is calculated using the formula:

\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean,
- \(\mu_0\) is the population mean under the null hypothesis (3.5 mg),
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

Substituting in the values:

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

Calculating the denominator:

\[
1.4 / \sqrt{8} \approx 1.4 / 2.828 \approx 0.495
\]

Now we compute the t-statistic:

\[
t = \frac{4.2 - 3.5}{0.495} \approx \frac{0.7}{0.495} \approx 1.414
\]

Thus, the value of the test statistic is approximately **1.414**.