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 not exceed 3.5 milligram? Use a 0.01 level of significance and assume the distribution of nicotine contents to be normal. The null and alternative hypotheses are ?

Question

GPT-4o mini · Answer

To test the manufacturer's claim regarding the average nicotine content, we need to set up the null and alternative hypotheses.

Null Hypothesis (H0): The average nicotine content does not exceed 3.5 milligrams. Mathematically, this can be expressed as: \[ H_0: \mu \leq 3.5 \]
Alternative Hypothesis (H1): The average nicotine content exceeds 3.5 milligrams. This can be expressed as: \[ H_1: \mu > 3.5 \]

We can now proceed to perform a statistical test to see if we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis at the 0.01 significance level.

Steps for the Hypothesis Test:

Sample Size (n): 8
Sample Mean (\(\bar{x}\)): 4.2 mg
Sample Standard Deviation (s): 1.4 mg
Significance Level (\(\alpha\)): 0.01

We'll perform a one-sample t-test because the sample size is small (n < 30), and we assume the normality of the distribution.

Test Statistic Calculation:

The t statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where:

\(\bar{x}\) = sample mean
\(\mu_0\) = population mean under the null hypothesis (3.5 mg)
\(s\) = sample standard deviation
\(n\) = sample size

Substituting in the values: \[ t = \frac{4.2 - 3.5}{1.4 / \sqrt{8}} = \frac{0.7}{1.4 / 2.8284} = \frac{0.7}{0.4949} \approx 1.415 \]

Critical Value:

Next, we determine the critical value from the t-distribution table for a one-tailed test with \(n - 1 = 7\) degrees of freedom at \(\alpha = 0.01\).

Using a t-table or calculator, the critical value for \(df = 7\) at \(\alpha = 0.01\) is approximately 2.998.

Decision Rule:

If the calculated t value exceeds the critical t value, we reject the null hypothesis.
If the calculated t value does not exceed the critical t value, we fail to reject the null hypothesis.

Conclusion:

Calculated t value: approximately 1.415
Critical t value: approximately 2.998

Since \(1.415 < 2.998\), we fail to reject the null hypothesis.

Final Statement:

There is not enough evidence at the 0.01 significance level to conclude that the average nicotine content of this brand exceeds 3.5 milligrams, supporting the manufacturer's claim.