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Homework: CH. 9 scored o of 10 pts 10 of 10 complete Instructor-created question

ID: 3207241 • Letter: H

Question

Homework: CH. 9 scored o of 10 pts 10 of 10 complete Instructor-created question HW Score: 18.80%, 1888 of 100 pts One male and one dam rat pup experiment the opposite end, accomplhed three was repeated unil each of a a 0.06. Comment ldentry the rejeasonregion for testing hypotheses with a-oos selectee comeet choice below and in answertex compit your choice. (Round to three deoma places an needed) Caloulate the test statistic t. (Round to three decimal places as needed.) Give the appropriate oonclusion forthe test in the rejection region,oondude that te mean number ofswims neaired for male pups T the mean number requined for female pups H. Since the test statistic c Do not reject Iswers and then click Check Answer

Explanation / Answer

Solution:

The null and alternative hypothesis for the given scenario of testing of hypothesis is given as below:

Here, we want to test whether there is any significant difference between the two population means or not. We have to use the t test for difference between two population means. We have to use two tailed test for checking this claim. Hypotheses are given as below:

Null hypothesis: H0: µd = 0

Alternative hypothesis: Ha: µd 0

This is a two tailed test.

So, correct answer is B.

Now, we have to find rejection region for this test.

Degrees of freedom = n – 1 = 8 – 1 = 7

Level of significance = alpha = 0.05

Critical values = -2.3646 and 2.3646

Rejection region is |t| > 2.3646

Answer: A. |t| > 2.365

Now, we have to find the test statistic value. The formula for test statistic is given as below:

Test statistic = t = Dbar / [Sd/sqrt(n)]

The calculation table is given as below:

Male

Female

Di

(Di – Dbar)^2

3

8

-5

33.0625

3

3

0

0.5625

7

7

0

0.5625

6

3

3

5.0625

7

6

1

0.0625

4

3

1

0.0625

9

3

6

27.5625

3

3

0

0.5625

Dbar = 0.75

Sd = 3.1053

Test statistic = t = 0.75/[3.1053/sqrt(8)

Test statistic = t = 0.683

p-value = 0.5165

P-value > Alpha value,

so we do not reject the null hypothesis H0 since the test statistic is not in the rejection region, conclude that the mean number of male pups is same as the mean number of female pups.

Male

Female

Di

(Di – Dbar)^2

3

8

-5

33.0625

3

3

0

0.5625

7

7

0

0.5625

6

3

3

5.0625

7

6

1

0.0625

4

3

1

0.0625

9

3

6

27.5625

3

3

0

0.5625