To determine if there is sufficient evidence to believe that the mean lifetime of mice on the new diet is more than 40 months, we can conduct a hypothesis test.
Step 1: Set up the hypotheses
- Null hypothesis (\(H_0\)): \(\mu \leq 40\) (The mean lifetime of the mice on the new diet is 40 months or less)
- Alternative hypothesis (\(H_a\)): \(\mu > 40\) (The mean lifetime of the mice on the new diet is greater than 40 months)
Step 2: Choose the significance level
Here, the significance level (\(\alpha\)) is given as 0.01.
Step 3: Determine the critical value
Since this is a one-tailed test (we are checking if the mean is greater), we will find the critical z-value for an upper-tail test at the 0.01 significance level.
Using a z-table or standard normal distribution calculator:
- The critical value for a one-tailed test at the 0.01 level of significance is \(z_{0.01} = 2.326\).
Step 4: Calculate the test statistic
We can calculate the test statistic using the formula: \[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \] Where:
- \(\bar{x} = 42\) (sample mean)
- \(\mu_0 = 40\) (hypothesized population mean)
- \(\sigma = 6\) (sample standard deviation)
- \(n = 65\) (sample size)
Plugging in the values: \[ z = \frac{42 - 40}{\frac{6}{\sqrt{65}}} \] First, calculate \(\frac{6}{\sqrt{65}}\): \[ \sqrt{65} \approx 8.0623 \Rightarrow \frac{6}{\sqrt{65}} \approx \frac{6}{8.0623} \approx 0.743 \] Now substitute back into the z-score formula: \[ z \approx \frac{2}{0.743} \approx 2.692 \]
Step 5: Decision
We compare the calculated \(z\) value to the critical value:
- Calculated \(z = 2.692\)
- Critical value \(z_{0.01} = 2.326\)
Since \(2.692 > 2.326\), we reject the null hypothesis.
Conclusion
There is sufficient evidence at the 0.01 significance level to conclude that the mean lifetime of the mice on the diet is greater than 40 months.
Answer
The critical value for this test is 2.326.