Question

In: Math

-Identify why you choose to perform the statistical test (Sign test, Wilcoxon test, Kruskal-Wallis test). -Identify...

-Identify why you choose to perform the statistical test (Sign test, Wilcoxon test, Kruskal-Wallis test).

-Identify the null hypothesis, Ho, and the alternative hypothesis, Ha.

-Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.

-Find the critical value(s) and identify the rejection region(s).

-Find the appropriate standardized test statistic. If convenient, use technology.

-Decide whether to reject or fail to reject the null hypothesis.

-Interpret the decision in the context of the original claim.

A weight-lifting coach claims that weight-lifters can increase their strength by taking vitamin E. To test the theory, the coach randomly selects 9 athletes and gives them a strength test using a bench press. Thirty days later, after regular training supplemented by vitamin E, they are tested again. The results are listed below. Use the Wilcoxon signed-rank test to test the claim that the vitamin E supplement is effective in increasing athletes' strength. Use α = 0.05.

Athlete

1

2

3

4

5

6

7

8

9

Before

185

241

251

187

216

210

204

219

183

After

195

246

251

185

223

225

209

214

188

Solutions

Expert Solution

We will use the Wilkoxon signed-rank test.

The paired samples Wilcoxon test (also known as Wilcoxon signed-rank test) is a non-parametric alternative to paired t-test used to compare paired data. It’s used when your data are not normally distributed.

Null hypothesis is: H0: di=0, where di= difference in the paired value xi , yi.

Alternate hypothesisis: H1: di : 0 (di not equal to 0)

The alternative hypothesis is always two tailed or two sided in case of wilcoxon signed rank test.

Below, i am attaching an image of the solution of the test using R software. i am including all the steps and commands so that it will be easy for u to run the program.

Here, V correspond to the sum of ranks assigned. P value is 0.06532, which is greater than the given value of alpha=0.05. Therefore, we accept the null hypothesis at 0.05 level of significance.

I have tried to answer as much as possible. Hope this helps. All the best..!


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