Question

Please write the full derivation of the answers. Thank you!

Q4) 2 points A researcher in the University of Toronto claims that the average life span of mice can be extended by as much as 8 months when the calories in their diet are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a random sample of 10 mice is fed a normal diet and has an average life span of 32.1 months with a standard deviation of 3.2 months, while a random sample of 15 mice is fed the restricted diet and has an average life span of 37.6 months with a standard deviation of 2.8 months. Test the hypothesis, at the 0.05 level of significance, that the average life span of mice on this restricted diet is increased by 8 months against the alternative that the increase is less than 8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal with equal variances.

Answer 1

$H_{0}: \mu1 - \mu2 \geq 8$

$H_{1}: \mu1 - \mu2 < 8$

The pooled variance (s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2) = (14 * (2.8)^2 + 9 * (3.2)^2)/(15 + 10 - 2) = 8.7791

The test statistic is

$t = \frac{(\bar x1 - \bar x2) - (\mu1 - \mu2)}{\sqrt{\frac{s_{p}^2}{n1} + \frac{s_{p}^2}{n2}}}$

  $= \frac{(37.6 - 32.1) - 8}{\sqrt{\frac{8.7791}{15} + \frac{8.7791}{10}}}$

df = 15 + 10 - 2 = 23

At alpha = 0.05, the critical value is -t_{0.05, 23} = -1.714

Since the test statistic value is less than the critical value (-2.07 < -1.714), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that the average life span of mice can be extended by as much as 8 months when the calorie in their diet are reduced by approximately 40% from the time they are weaned.

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