Page 78 - Basic Statistics

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Therefore t-actual < - t0.05;41, which is - 8.48 < -1.684 then H0 is rejected at the 5

percent significance level. It was concluded that the mean test scores of the

direct learning lower than the mean test scores of the cooperative learning.

Solving using computerized:

The data in columns 2 and 3 of Table 2 stored in the S-plus, called TPS.

Its data in the form of a matrix and has the number of row and column 25 and 2

respectively.

> var.test(TPS[,1],TPS[,2],alternative="two.sided", conf.level=.95)

F test for variance equality

data: TPS[, 1] and TPS[, 2]
F = 0.4245, num df = 24, denom df = 24, p-value = 0.0407
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.1870554 0.9632639

sample estimates:
variance of x variance of y
16.75 39.46

> t.test(TPS, TPS [, 2], alternative="two.sided", mu=0, paired=F,
+ var.equal=F, conf.level=.95)

Welch Modified Two-Sample t-Test

data: x: direct in TPS , and y: cooperative in TPS
t = -8.483, df = 41.264, p-value = 0
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-15.747645 -9.692355
sample estimates:
mean of x mean of y
64 76.72

> t.test(TPS [, 1], TPS [, 2], alternative="less", mu=0, paired=F,
+ var.equal=F, conf.level=.95)

Welch Modified Two-Sample t-Test

~~* CHAPTER 4 TESTING HYPOTHESIS *~~