# ent-measuresstudy, these individual differences can cause prob-lems. Identify the problems and briefly explain howthey are eliminated or reduced with a repeated-22. TOnedsmeasures study.21. In the Chapter Preview we described a studyshowing that students had more academic problemsfollowing nights with less than average sleepcompared to nights with more than average sleep(Gillen-O'Neel, Huynh, & Fuligni, 2013). Supposeo researcher is attempting to replicate this studyusing a sample of n 8 college freshmen. Eachstudent records the amount of study time and amountof sleep each night and reports the number ofacademic problems each day. The following datashow the results from the study.R.3 ANOVANdtorNumber of Academic Problems12Exees of HyFollowing Nightswith AboveFollowing Nightswith Belowe TestsAverage SleepStudentAverage Sleep1310АSuCy68ВSolt9 ProblerFocus612.1emp6alD4EProb1410F131153H lar363PROBLEMSa. Treat the data as if the scores are from an indepen-dent-measures study using two separate samples,each with n8 participants. Compute the pooledvariance, the estimated standard error for themean difference, and the independent-measurest statistic. Using a two-tailed test with o=there a significant difference between the two setsof scores?.05, isb. Now assume that the data are from a repeated-8measures study using the same sample of n =participants in both treatment conditions. Computethe variance for the sample of difference scores,the estimated standard error for the mean differ-ence and the repeated-measures t statistic. Using atwo-tailed test with a .05, is there a significantdifference between the two sets of scores? (Youshould find that the repeated-measures design sub-stantially reduces the variance and increases thelikelihood of rejecting Ho.)2. The previous problem demonstrates that remov-ing individual differences can substantially reducevariance and lower the standard error. However, thisbenefit only occurs if the individual differences areconsistent across treatment conditions. In problem 2.1

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Number 21

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Step 1

Independent-measures t test:

Denote μ1, μ2 as the true average number of academic problems on nights after below average sleep, and under above average sleep, respectively. The null and alternative hypotheses are as follows:

H0: μ1 = μ2 versus H1: μ1 ≠ μ2.

The two-sample t test is to be conducted, at level of significance α = 0.05.

We have entered the data in an Excel sheet, with the data on below average sleep in column A, and that on above average sleep in column B. The first row has been used for labels.

The respective sample means are, 1 = 9 and 2 = 7 [Using Excel formulae: =AVERAGE(A2:A9) and =AVERAGE(B2:B9) respectively].

The sample variances are respectively, s12 = 14.28571, s22 = 9.71429 [Using Excel formulae: =VAR.S(A2:A9) and =VAR.S(B2:B9) respectively].

Here, n1 = n2 = 8

The degrees of freedom is, df = n1 + n2 – 2 = 8 + 8 – 2 = 14.

The pooled variance is:

s2 = [(n1 – 1) s12 + (n2 – 1) s22] / (n1 + n2 – 2)

= [(7 × 14.28571) + (7 × 9.71429)] / 14

= 12.

Thus, pooled variance = 12.

The standard error for the mean difference is:

se = s√ (1/n1 + 1/n2)

= √[s2(1/n1 + 1/n2)]

= √ [12 (1/8 + 1/8)

= 1.7321.

Thus, standard error of mean difference = 1.7321.

The test statistic is found as follow...

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