Week 5 - Bobby Illig 9.12 9.16 Question 9.18 9.26 11.20 The...

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Bobby Illig9.129.16
Question 9.18
9.26
11.20The mean of students in the front - mean of the student in the back = (82+ 79.5+ 71.5+ 72)/4 - (78+77.41 +76+ 71.3)/411.22Let the first group be 'Liberals'.
Second group is 'Conservative'Using MINITAB software, the calculated are:a.Variance of first groups s2= 0.667Variance of second group s2=1.100b.Degrees of freedom of X(liberals) = (n1- 1) =3Degrees of freedom of Y(conservative) = (n2-1) =5Degrees of freedom of total 10 — 2 = 8c.Critical value oft is 2.3060 at 0.05 leveld.Pooled variance is 0.937411e.Standard error of first group is 0.408 Standard error of second group is 0.428f.F variance is 0.61g. t statisticIt = —2.40h. The 95% confidence interval is, (-2.941,-0.0599. Cohen's value:
11.26a) A total of 73 people were studied, 40 in on group and 33 in the other group. The test statistic iscalculated as 2.13 The critical value fora two-tailed test with a p — level of 0.05 is ±-1.96 As teststatistics value (2.13) is greater than critical value (1.96), So we reject the null hypothesis as test
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Term
Spring
Professor
Blaydes,KathyJ.
Tags
Statistics, Null hypothesis, Statistical hypothesis testing, critical value, critical value fora, Bobby Illig

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