Question

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 not equal to the variance in group B

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. Report the P value, rounded to 3 decimal places, in the box below. Do not use the rounded value for F when calculating the P value. Use the F 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.

Here is the data:

A	B
42	48
8	131
27	24
40	79
9	91
67	143
53	27
35	86
73	54
80	12
35	72
38	115
38	31
53	53
46	125
61	109
47	99
6	25
28	130
43	104

Answer 1

THE HYPOTHESIS HERE IS ALETRNATE HYPOTHESIS : VARIANCE OF GROUP A IS NOT EQUAL TO VARIANCE OF GROUP B.

F-Test Two-Sample for Variances
	A	B
Mean	41.45	77.9
Variance	408.6815789	1718.726316
Observations	20	20
df	19	19
F	0.23778165
P(F<=f) one-tail	0.001479554
F Critical one-tail	0.461201089

NULL HYPOTHESIS IS: TWO SAMPLKE VARIANCES ARE EQUAL

since calculated value of F(0.23778165) is is less than critical value of F(0.46120),so we accept the null hypothesis .i.e. their is no significant difference between the population variances.

2)= I suggest students t test if we are testing difference between two population means

t-Test: Two-Sample Assuming Equal Variances
	A	B
Mean	41.45	77.9
Variance	408.6815789	1718.726316
Observations	20	20
Pooled Variance	1063.703947
Hypothesized Mean Difference	0
df	38
t Stat	-3.53416765
P(T<=t) one-tail	0.000546898
t Critical one-tail	1.68595446
P(T<=t) two-tail	0.001093796
t Critical two-tail	2.024394164

since calculated value of t is less than critical value of t for both one tail and two tail so we accept the null hypothesis i.e. their is no significant difference between means of two groups