Question

Here is the data set concerning treatment groups A and B.

57	70
66	27
65	15
40	86
64	49
11	16
34	57
61	29
26	73
45	71
30	98
43	46
64	88
19	22
14	53
14	59
11	94
34	18
68	68
39	62

Before performing a t-test to determine if the two groups are different, researchers must first ascertain if the variance in the two groups is the same using an F test.

Indicate whether the hypothesis below is the null or alternate hypothesis for the F test.

"The variance in group A is equal to the variance in group B" _____________? Null/Alternate  

Calculate the F statistic for this data set and report it in the box below, rounded to two decimal places. ____________?

Use the formula =(1-(f.dist(F, df₁,df₂,true) in Excel to calculate the P value for the F statistic, and report it in the box below. Do not use the rounded value for F when calculating the P value. Use the value in the spreadsheet that has not been rounded. __________?

Based on this p value, would you recommend using a Student's t test or a Welch's t test to determine if the two treatment groups have different means? Enter either Student's or Welch's in the box below. ____________?

Answer 1

Using R,

CODE:

a <- fread("57 70

66 27

65 15

40 86

64 49

11 16

34 57

61 29

26 73

45 71

30 98

43 46

64 88

19 22

14 53

14 59

11 94

34 18

68 68

39 62")

names(a) <- c("A","B")

var.test(a$A,a$B,alternative = "two.sided")

OUTPUT:

> a <- fread("57 70

+ 66 27

+ 65 15

+ 40 86

+ 64 49

+ 11 16

+ 34 57

+ 61 29

+ 26 73

+ 45 71

+ 30 98

+ 43 46

+ 64 88

+ 19 22

+ 14 53

+ 14 59

+ 11 94

+ 34 18

+ 68 68

+ 39 62")

> names(a) <- c("A","B")

> var.test(a$A,a$B,alternative = "two.sided")

F test to compare two variances

data: a$A and a$B

F = 0.57577, num df = 19, denom df = 19, p-value =

0.238

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

0.2278986 1.4546665

sample estimates:

ratio of variances

0.5757747

"The variance in group A is equal to the variance in group B" : Null hypothesis.

The value of the F-statistic is 0.58.

P-value is 0.238.

Here p-value is > 0.05. So, we can't reject the null hypothesis that variances are equal at 5% level of significance.

So, as the variances are equal, we use Student's t-test to determine if the two treatment groups have different means.

Student's t-test using R:

CODE:

t.test(a$A,a$B,alternative = "two.sided", var.equal = T)

OUTPUT:

> t.test(a$A,a$B,alternative = "two.sided", var.equal = T)

Two Sample t-test

data: a$A and a$B

t = -1.9691, df = 38, p-value = 0.05626

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-30.0154893 0.4154893

sample estimates:

mean of x mean of y

40.25 55.05

So, the P-value of the t-test is 0.05626 > 0.05. So, we can't reject the null hypothesis that the two means are equal at 5% level of significance,

So, we can conclude that the means are equal.