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Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according...

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)

Diet A 5 13 9 8 10 14 5 8 7 5

Diet B 15 10 11 13 16 11 20 11 10 13

(a) Find t. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer correct to four decimal places.)

Is there a way to do this problem on a TI-84 Plus calculator? If so can you please break down the steps in getting the answers on the calculator? Thank you!

Solutions

Expert Solution

Solution:-

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

Null hypothesis: uB - uA< 0
Alternative hypothesis: uB - uA > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 1.4314
DF = 18

t = [ (x1 - x2) - d ] / SE

t = 3.21

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 3.21.

Therefore, the P-value in this analysis is 0.0024.

Interpret results. Since the P-value (0.0024) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the mean weight gained on diet B was greater than the mean weight gained on diet A.


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