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.

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.
Transcript text: 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 $\alpha=0.01$, 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. \[ \begin{array}{l} H_{0}: \square \text { (Choose one) } \mathbf{V} \\ H_{1}: \square \text { (Choose one) } \mathbf{\nabla} \end{array} \] $\square$ (Choose one) test.
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Solution

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Solution Steps

Step 1: State the Hypotheses

We define the null and alternative hypotheses as follows: H0:μ=33(The mean number of cigarettes smoked per day is 33) H_0: \mu = 33 \quad \text{(The mean number of cigarettes smoked per day is 33)} H1:μ<33(The mean number of cigarettes smoked per day is less than 33) H_1: \mu < 33 \quad \text{(The mean number of cigarettes smoked per day is less than 33)}

Step 2: Calculate the Standard Error

The standard error SESE is calculated using the formula: SE=σn=2.510=0.7906 SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{10}} = 0.7906

Step 3: Calculate the Test Statistic

The test statistic ttestt_{\text{test}} is calculated using the formula: ttest=xˉμ0SE=27330.7906=7.5895 t_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{27 - 33}{0.7906} = -7.5895

Step 4: Determine the P-value

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

Step 5: Compare the P-value with the Significance Level

We compare the P-value with the significance level α=0.01\alpha = 0.01: P<α0.0<0.01 P < \alpha \quad \Rightarrow \quad 0.0 < 0.01

Step 6: Make a Decision

Since the P-value is less than the significance level, we reject the null hypothesis H0H_0.

Step 7: Conclusion

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

Final Answer

The hypotheses are: H0:μ=33(The mean number of cigarettes smoked per day is 33) H_{0}: \mu = 33 \quad \text{(The mean number of cigarettes smoked per day is 33)} H1:μ<33(The mean number of cigarettes smoked per day is less than 33) H_{1}: \mu < 33 \quad \text{(The mean number of cigarettes smoked per day is less than 33)} The claim is supported. Thus, there is enough evidence to conclude that the mean number of cigarettes smoked per day is less than 33.

There is enough evidence to support the claim.\boxed{\text{There is enough evidence to support the claim.}}

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