To determine if there is evidence to suggest that the mean lifetime of mice on the diet is smaller than 40 months, we need to set up our null and alternative hypotheses, perform a hypothesis test, and calculate the relevant statistics.
Null and Alternative Hypotheses:
- Null Hypothesis (H0): The mean lifetime of mice on the diet is equal to 40 months. (H0: μ = 40)
- Alternative Hypothesis (H1): The mean lifetime of mice on the diet is less than 40 months. (H1: μ < 40)
Given Data:
- Sample mean (\( \bar{x} \)) = 38 months
- Sample standard deviation (s) = 5.8 months
- Sample size (n) = 64 mice
- Significance level (α) = 0.05
Test Statistic:
Since the sample size is large \( (n > 30) \), we can use the z-test for the hypothesis test. The test statistic (z) is calculated as follows:
\[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Where:
- \( \mu_0 \) = 40 months (mean under the null hypothesis)
- \( \bar{x} \) = 38 months (sample mean)
- \( s \) = 5.8 months (standard deviation)
- \( n \) = 64 (sample size)
Calculating the test statistic:
- Calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{5.8}{\sqrt{64}} = \frac{5.8}{8} = 0.725 \]
- Then, calculate the z-score:
\[ z = \frac{38 - 40}{0.725} = \frac{-2}{0.725} \approx -2.76 \]
Finding the p-value:
Now, we will find the p-value for the calculated z-score. For a z-score of -2.76, we can use a standard normal distribution table or a calculator to find the p-value.
The p-value corresponds to \( P(Z < -2.76) \). Using a z-table or calculator, we find that:
\[ p \text{-value} \approx 0.0029 \]
Decision Rule:
- If the p-value is less than the significance level (α = 0.05), we reject the null hypothesis.
- If the p-value is greater than or equal to α, we fail to reject the null hypothesis.
Conclusion:
Since \( p \text{-value} \approx 0.0029 < 0.05 \), we reject the null hypothesis.
There is significant evidence to suggest that the mean lifetime of the mice on the vitamin and protein-rich diet is smaller than 40 months.
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