Question

Question 1:Professor Handy measured the time in seconds required to catch a falling meter stick for 12 randomly selected students’ dominant hand and nondominant hand. The Minitab Express file contains these measurements. Professor Handy claims that the reaction time in an individual’s dominant hand is less than the reaction time in their nondominant hand. Assuming that the differences follow a normal distribution, test the claim at the 5% significance level.

Question 2:The New England Patriots. The 2017 roster of the New England Patriots, winners of the 2017 NFL Super Bowl included 12 defensive linemen and 9 offensive linemen. The Minitab Express file for this problem contains the weights in pounds of the offensive and defensive linemen. Use this data set to test the claim that the defensive linesmen weigh less that the offensive linemen at the 5% level of significance.

Question 3: Stress and weight in rats. In a study of the effects of stress on weight in rats, 71 rats were randomly assigned to either a stressful environment or a control (nonstressful) environment. After 21 days, the change in weight (in grams) was determined for each rat. The table below summarizes the data on weight gain. Test the claim that stress effects weight. (Use a 10% significance level.)

Group	n	Sample mean	Sample Standard Dev.
Stress	20	26	13.4
No stress	51	32	14.2

Please provide all answers in the following format:

Step 1: State the null and alternative hypothesis. (Use “mu” for the symbol μ.)

Step 2: Calculate the test statistic.

Step 3: Find the p-value.

Step 4: State your conclusion. (Do not just say “Reject H0” or “Do not reject H0.” State the conclusion in the context of the problem.)

Answer 1

Solution:-

3)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Stress> u_{No Stress}
Alternative hypothesis: u_Stress < u_{No Stress}

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 3.6342
DF = 69

t = [ (x₁ - x₂) - d ] / SE

t = - 1.65

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of - 1.65.

Therefore, the P-value in this analysis is 0.052

Interpret results. Since the P-value (0.052) is less than the significance level (0.10), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that stress effects weight.