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Asked Nov 27, 2019

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Number 21

Step 1

**Independent-measures t test:**

Denote *μ*1, *μ*2 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: *μ*1 = *μ*2 versus H1: *μ*1 ≠ *μ*2.

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, *x̅*1 = 9 and *x̅*2 = 7 [Using Excel formulae: =AVERAGE(A2:A9) and =AVERAGE(B2:B9) respectively].

The sample variances are respectively, *s*12 = 14.28571, *s*22 = 9.71429 [Using Excel formulae: =VAR.S(A2:A9) and =VAR.S(B2:B9) respectively].

Here, *n*1 = *n*2 = 8

The degrees of freedom is, df = *n*1 + *n*2 – 2 = 8 + 8 – 2 = 14.

The pooled variance is:

*s*2 = [(*n*1 – 1)* s*12 + (*n*2 – 1)* s*22] / (*n*1 + *n*2 – 2)

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

= 12.

Thus, **pooled variance = 12**.

The standard error for the mean difference is:

*se* = *s*√ (1/*n*1 + 1/*n*2)

= √[*s*2(1/*n*1 + 1/*n*2)]

= √ [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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