Question

Forty students are selected to take part in an experiment testing the effect of espresso consumption on math test scores. The students are split into four groups of ten: a control group is given no espresso, another group is given a single espresso, another is given a double espresso, and the last is given a triple espresso. After the espresso is administrated, the students take a math test. The sample means and standard deviations of the results are: . • No espresso: mean = 49.0, SD = 20.9 One espresso: mean = 55.1, SD = 18.6 • Two espressos: mean = 64.2, SD = 16.9 • Three espressos: mean = 53.8, SD = 11.3 Assume that each group's scores are approximately normally distributed. Test the hypothesis that the mean scores for the four groups are the same at the a = 0.05 of significance. a. What is the F test statistic? Please round the answer to 1 decimal place. (A) reject the null b. Using either classical approach or P-value approach, you will hypothesis, or (B) fail to reject the null hypothesis. Please type either A or B.

Answer 1

Ans : 40 students are selected to take part in an experiment testing the effect of espresso consumption on math test score .

given that the students are split into 4 group of 10

given that

No Espresso : mean = 49 , standard deviation = 20.9

one Espresso : mean = 55.1 , standard deviation = 18.6

two Espresso : mean = 64.2 , standard deviation = 16.9

three Espresso : mean = 53.8 , standard deviation = 11.3

this can be summarized as :

Espresso	j	. 22	si
1	10	49	20.9
2	10	55.1	18.6
3	10	64.2	16.9
4	10	53.8	11.3

Then

r
_1. = 49 ,
r
_2. = 55.1 ,
r
_3. = 64.2 ,
r
4. =53.8

s₁ = 20.9 ,  s₂ = 18.6 , s₃ = 16.9 , s4 = 11.3

Suppose there are i treatment and j observation corresponding to each treatment

Degrees of freedom treatment = i-1

Degrees of freedom error = i*(j-1)

Degrees of freedom total = i*j -1

One way ANOVA :

Source	DF	SS	MS	F
Treatment	i-1	SSTreatment	SSTreatment /(i-1)	MSTreatment/MSError
Error	i*(j-1)	SSError	SSError /(i*(j-1))
Total	i*j-1	SSTotal

From the given problem

Degrees of freedom treatment = 4-1 = 3

Degrees of freedom error =4*(10 -1 ) = 36

Degrees of freedom total = 4*10-1 = 39

Grand mean :

X
.. =
* Χι.

49+ 55.1 + 64.2 + 53.8 4

.. = 55.525

X

To test the hypothesis that the mean scores for four groups are the same .

test for Hypothesis :

H₀ : $1596122980398_blob.png$ 1 = $1596122980523_blob.png$ 2 = $1596122980396_blob.png$ 3 Versus H₁ : $1596122980483_blob.png$ i $1596122980504_blob.png$ 0 , for at least one i = 1,2,3,

Test Statistic :

F= MS_Treatment / MS_Error

Now first calculate  Following

SS_Treatment = j * $1596123019887_blob.png$ (
22
_. -
X
..)^2

SS_Treatment =    10{(49 - 55.525)^2 + (55.1-55.525)^2+(64.2-55.525)^2 + (53.8 - 55.525 )^2}

SS_Treatment = 1209.875

MS_Treatment = 1209.875/ 3

MS_Treatment = 403.2917

SS_Error = (j-1)*$1596123019891_blob.png$ (S_i)^2

SS_Error = (10-1)*{20.9^2+18.6^2+16.9^2 + 11.3^2}

SS_Error = 10764.63

MS_Error = 10764.63/36

MS_Error = 299.0175

SS_Total = SS_Error + SS_Treatment

SS_Total = 11974.51

Test Statistic :

F= MS_Treatment / MS_Error

F =403.2917 / 299.0175

F= 1.3

The F test statistic = 1.3

p-value :

p-value = 0.28893

since p - value =0.28893 > 0.05 Then we fail to reject H0 at 0.05 significance level

b) using either classical approach or p -value approach , you will B

ANOVA :

Source	DF	SS	MS	F
Treatment	3	1209.875	403.2917	1.3
Error	36	10764.63	299.0175
Total	39	11974.51