91Ó°ÊÓ

A UCLA researcher claims that the average life span of mice can be extended by as much as 8 months when the calories in their food 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 are fed a normal diet and live an average life span of 32.1 months with a standard deviation of 3.2 months. while a random sample of 15 mice are fod the restricted diet and live 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.

Step by step solution

01

Null and Alternative Hypothesis

The null hypothesis \( H_0 \) is that the average lifespan increase due to the restricted diet is equal to 8 months. The alternative hypothesis \( H_1 \) is that the increase is less than 8 months. Mathematically, these are represented as: \( H_0: \mu_1 - \mu_2 = 8 \) and \( H_1: \mu_1 - \mu_2 < 8 \) where \( \mu_1 \) is the mean lifespan of mice with a restricted diet and \( \mu_2 \) is the mean lifespan of mice with a normal diet.

02

Calculate the Test Statistic

The formula for the test statistic in a two-sample hypothesis test for population means with equal variances is \( Z = \frac{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)} { \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} } \). Here \( \bar{x_1} = 37.6, \bar{x_2} = 32.1, s_1 = 2.8, s_2 = 3.2, n_1 = 15, n_2 = 10, \mu_1 - \mu_2 = 8 \). Substituting these values into the formula gives the test statistic value.

03

Find the Critical Value

The level of significance is given as 0.05. For a one-tailed test with this level of significance, the critical value of Z is -1.645 (negative because it is a 'less-than' test). This value can be found using standard normal distribution tables.

04

Make your final decision

Compare the test statistic to the critical value. If the test statistic is less than the critical value, reject the null hypothesis. If not, do not reject it. This is the final decision in the hypothesis test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null and Alternative Hypothesis

In hypothesis testing, the null hypothesis (\( H_0 \)) is a statement of no effect or no difference and serves as a baseline for comparison. For our UCLA researcher's claim, the null hypothesis suggests that the average life span increase due to the restricted diet is exactly 8 months. In contrast, the alternative hypothesis (\( H_1 \) or \( H_a \) represents that the average life span increase is different from the null hypothesis' prediction—in this case, less than 8 months.

Formulating these hypotheses is critical as they define the direction and scope of the study. It's crucial to state them correctly before any testing since they guide the entire testing process and help to avoid misinterpretation of results.

Test Statistic Calculation

The test statistic is a normalized value calculated from sample data during a hypothesis test. It quantifies how far the sample statistic deviates from the null hypothesis. To calculate the test statistic, we use the formula for a two-sample t-test under the assumption of equal variances.

In our exercise, the test statistic is given by \[ Z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)} {\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. The subtraction of the hypothesized difference (\(\mu_1 - \mu_2\)) ensures we are measuring the distance from the null hypothesis' claim.

Critical Value Determination

The critical value serves as a threshold to decide whether to reject the null hypothesis. It's determined by the significance level (\(\alpha\)), which is the probability of rejecting the null hypothesis when it is true (Type I error).

In a one-tailed test, like the one we're discussing, critical values are one-sided and indicate the cutoff point on only one end of the distribution. Using normal distribution tables or a software, we find the critical value corresponding to the significance level. For a level of significance at 0.05 in a 'less than' direction, the critical value is \(Z_{\alpha} = -1.645\). This number effectively creates a boundary; if our test statistic falls to the left of \(Z_{\alpha}\), the null hypothesis is rejected.

One-Tailed Test

The one-tailed test is designed to determine if there a significant difference in a specific direction (either greater than or less than). The UCLA researcher's hypothesis, which predicts a specific direction of change (less than 8 months), requires a one-tailed test.

To execute the test, we compare the calculated test statistic against the critical value. If the test statistic is in the critical region (for our case, less than \(Z_{\alpha}\)), we reject the null hypothesis, concluding that there is enough evidence to support the researcher's claim that the increase in life span is less than 8 months. Conversely, if the test statistic is greater than the critical value, we fail to reject the null hypothesis, implying that the study did not find adequate evidence to support the claim.