To test the claim that the mean waiting time for bus number 14 during peak hours is less than 10 minutes, we need to establish the null and alternative hypotheses, calculate the test statistic, and then interpret the P-value to make a conclusion.
Step 1: Set up the hypotheses
- Null hypothesis (\(H_0\)): \(\mu \geq 10\) (the mean waiting time is greater than or equal to 10 minutes)
- Alternative hypothesis (\(H_a\)): \(\mu < 10\) (the mean waiting time is less than 10 minutes)
Step 2: Gather the sample information
- Sample size (\(n\)) = 18
- Sample mean (\(\bar{x}\)) = 7.9 minutes
- Sample standard deviation (\(s\)) = 1.5 minutes
- Significance level (\(\alpha\)) = 0.01
Step 3: Calculate the test statistic
Since the population is normally distributed and we are using a sample standard deviation, we will use the t-distribution.
The formula for the t-statistic is:
\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
Where:
- \(\mu_0\) is the population mean under the null hypothesis (10 minutes).
Substituting the values:
\[ t = \frac{7.9 - 10}{1.5/\sqrt{18}} = \frac{-2.1}{1.5/\sqrt{18}} = \frac{-2.1}{1.5/4.24264} = \frac{-2.1}{0.35355} \approx -5.940 \]
Step 4: Find the critical value and P-value
We will find the P-value for the t-statistic calculated. The degrees of freedom \(df = n - 1 = 18 - 1 = 17\).
Using a t-distribution table or calculator, we can find the P-value for \(t = -5.940\) with 17 degrees of freedom. The P-value is very small (less than 0.001), indicating strong evidence against the null hypothesis.
Step 5: Conclusion
Since the P-value is less than the significance level \(\alpha = 0.01\), we reject the null hypothesis.
Thus, we have sufficient evidence to support the claim that the mean waiting time is less than 10 minutes.
Final Answer
The correct choice based on the information and calculations is:
B) Test statistic: t = –5.940. There is sufficient evidence to support the claim that the mean is less than 10 minutes.