Question

The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a random sample of 10 cigarettes of this brand. The sample yields an average of 1.55 mg of nicotine. Conduct a test using a significance level of α=0.01.

Answer 1

To conduct the test, we will use a one-sample t-test. The null hypothesis (H0) is that the mean nicotine content is 1.5 mg, and the alternative hypothesis (H1) is that the mean nicotine content is higher than 1.5 mg. The significance level (α) is 0.01.

Step-by-step explanation:

To conduct the test, we will use a one-sample t-test. The null hypothesis (H0) is that the mean nicotine content is 1.5 mg, and the alternative hypothesis (H1) is that the mean nicotine content is higher than 1.5 mg. The significance level (α) is 0.01.

First, calculate the t-value using the formula: t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size)). In this case, the sample mean is 1.55 mg, the population mean is 1.5 mg, the sample standard deviation is 0.1 mg, and the sample size is 10. Plugging in these values, we get t = (1.55 - 1.5) / (0.1 / sqrt(10)) = 1.5811.

Next, compare the t-value with the critical t-value. The degrees of freedom (df) for this test is the sample size minus 1, which is 10 - 1 = 9. For a one-tailed test with a significance level of 0.01 and df = 9, the critical t-value is approximately 2.8214 (obtained from a t-table or calculator).

Since the calculated t-value (1.5811) is less than the critical t-value (2.8214), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean nicotine content is higher than 1.5 mg.