A researcher assesses the level of a particular hormone in the blood in five pat

Question

A researcher assesses the level of a particular hormone in the blood in five patients before and after they begin taking a hormone treatment program. Results for the five are shown to the right. Using he 0.05 significance level, was there a significant change in the level of this hormone? Complete parts (a) through (c). 0.20 0.18 0.22 0.22 c 0.23 0.22 D 0.18 0.16 E 0.20 0.18 The cutoff sample score(s) is/are (Use a comma to separate answers as needed. Round to three decimal places as needed.) Determine the sample's score on the comparison distribution. Use After- Before to compute the differences. The sample's score on the comparison distribution is (Round to three decimal places as needed.) Decide whether to reject the null hypothesis. Choose the correct answer below A. Reject the null hypothesis because the sample's score is more extreme than the cutoff sample score(s). O B. Reject the null hypothesis because the sample's score is less extreme than the cutoff sample score(s). c. Fail to reject the null hypothesis because the sample's score is less extreme than the cutoff sample score(s). O D. Fail to reject the null hypothesis because the sample's score is more extreme than the cutoff sample score(s). distribution. Label the sample score and cutoff sample score(s) for the test. Select the correct choice below and fill in the answer box to

Explanation / Answer

The cutoff sample scores are : -t0.025;4 , t0.025;4
                                                     = -2.776 , 2.776

Sample score : n . mean(d) / s.d(d), where d= before-after
                        = 2.714

Conclusion :
Option(C) Fail to reject the null hypothesis since the sample score is less extreme than the cutoff sample score.

R codes :

> before=c(0.2,0.22,0.23,0.18,0.2)

> after=c(0.18,0.22,0.22,0.16,0.16)

> t.test(before,after,conf.level=0.95,paired=T)

        Paired t-test

data: before and after

t = 2.7136, df = 4, p-value = 0.05334

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.0004168533 0.0364168533

sample estimates:

mean of the differences

                  0.018

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