Questions: Cigarette Smoking A researcher found that a cigarette smoker smokes on average 33 cigarettes a day. She feels that this average is too high. She selected a random sample of 10 smokers and found that the mean number of cigarettes they smoked per day was 27. The sample standard deviation was 2.5. At α=0.01, is there enough evidence to support her claim? Assume that the population is approximately normally distributed. Use the critical value method and tables. Part 1 of 5 (a) State the hypotheses and identify the claim. H0: (Choose one) V H1: (Choose one) ∇ (Choose one) test.

Solution Steps

We define the null and alternative hypotheses as follows: \[ H_0: \mu = 33 \quad \text{(The mean number of cigarettes smoked per day is 33)} \] \[ H_1: \mu < 33 \quad \text{(The mean number of cigarettes smoked per day is less than 33)} \]

The standard error \(SE\) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{10}} = 0.7906 \]

The test statistic \(t_{\text{test}}\) is calculated using the formula: \[ t_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{27 - 33}{0.7906} = -7.5895 \]

For a left-tailed test, the P-value is calculated as: \[ P = T(z) = 0.0 \]

We compare the P-value with the significance level \(\alpha = 0.01\): \[ P < \alpha \quad \Rightarrow \quad 0.0 < 0.01 \]

Since the P-value is less than the significance level, we reject the null hypothesis \(H_0\).

There is enough evidence to support the claim that the mean number of cigarettes smoked per day is less than 33.

Final Answer