Compare the population means for the two distribution means below: a <- p(6, 7,

Question

Compare the population means for the two distribution means below:

a <- p(6, 7, 5, 8, 4, 3, 3, 9, 10)

b <- p(6, 5, 3, 4, 0, 0, 7, 9, 8)

A) Assuming that you want a 0.05 significance level in testing the null hypothesis, should we reject?

B) If 'means are equal' compare the means of the population from a. b samples using a paired t-test. Assume a Significance level of 0.05

C) If 'means are equal' compare the means of the population from a, b samples using a non-paired t-test. Assume a Siginificance level of 0.05

D) Why are the calculations carried above in questions not suitable for the data in a, b?

Explanation / Answer

Compare the population means for the two distribution means

First we ferform compare mean but before proceeding with the t-test, it is necessary to evaluate the sample variances of the two groups, using a Fisher’s F-test to verify the variance ar equal or not .

A) We use F-test to compare variances.

Ho= variance of a and b are equal

Vs

Ha : variance of a and b are not equal

> a=c(6,7,5,8,4,3,3,9,10)
> b=c(6,5,3,4,0,0,7,9,8)
> var.test(a,b)

F test to compare two variances

data: a and b
F = 0.62963, num df = 8, denom df = 8, p-value = 0.5277
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.1420241 2.7913118
sample estimates:
ratio of variances
0.6296296

Decision : we can see pvalue is greter than 0.05 so we do not reject null hypothesis .that means variances are equal

Now

Comapre means using t test eual variance .

Ho=means of a and b are equal

vs

Ha=means of a and b are not equal

> t.test(a,b,var.equal=TRUE,paired=FALSE)

Two Sample t-test

data: a and b
t = 1.0476, df = 16, p-value = 0.3104
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.478593 4.367481
sample estimates:
mean of x mean of y
6.111111 4.666667

Decision : we can see pvalue is greter than 0.05 so we do not reject null hypothesis .that means the mean of a and b are equal

B)

Ho:true mean differnce is equal / dependent.

Vs

Ha: true mean differnce is not equal / independent.

> t.test(a,b,var.equal=TRUE,paired=TRUE)

Paired t-test

data: a and b
t = 1.7295, df = 8, p-value = 0.122
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.4814913 3.3703802
sample estimates:
mean of the differences
1.444444

Decision : we can see pvalue is greter than 0.05 so we do not reject null hypothesis .that means the true mean difference is not equal

C)

Ho :true difference in means is equal to 0

vs

Ha:true difference in means is not equal to 0

> t.test(a,b)

Welch Two Sample t-test

data: a and b
t = 1.0476, df = 15.214, p-value = 0.3112
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.490912 4.379800
sample estimates:
mean of x mean of y
6.111111 4.666667

Decision : we can see pvalue is greter than 0.05 so we do not reject null hypothesis .that means the true mean difference is not equal 0

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Best of Luck :)

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