Question

QUESTION 5 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 Ftest statistic? Please round the answer to 1 decimal place. b. Using either classical approach or P-value approach, you will the null hypothesis. Please type either A or B. (A) reject the null hypothesis, or (B) fail to reject

Answer 1

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

given that

N = 40

For No espresso :

mean = 1. = 49

X

standard deviation = s1 = 20.9

For one espresso :

2. = 55.1

X

standard deviation = s2 = 18.6

For two espresso :

3. = 64.2

X

standard deviation = s3 =16.9

For three espresso :

4. = 53.8

X

standard deviation = s1 = 11.3

claim that test the hypothesis that the mean scores for the four groups are the same at the α = 0.05 of significance

test of Hypothesis :

H₀ : µ1 = µ2 =µ3=µ4 versus H₁ : at least one µi ≠ 0 for i = 1,2,3,4

Test Statistic :

F= MS_treatment / MS_Error

One way ANOVA :

Source	DF	Sum of Squares	Mean square (MS)	F
Treatment	k-1	SSTreatment	MSTreatment	MSTreatment/MSError
Error	N-k	SSError	MSError
Total	N-1	SStotal

Degrees of Freedom of treatment : DF = k − 1 , where k is the number of groups

df= 4-1 = 3

Degrees of Freedom Error :  DF = N − k , where N is the total number of subjects

df = 40 - 4 = 36

Total Degrees of Freedom: DF = N − 1

df = 40-1 = 39

The students are split into four groups of ten

so n_i = 10 for i = 1 ,2,3,4

χ..= Σχι. k

= 222.1/4

=55.52

k SStreatment ni(x7.– 2..)? i=1

SS_Treatment = 10(49-55.52)^2 + 10*(55.1 - 55.52)^2 + 10*(64.2- 55.52)^2 + 10*(53.8-55.52)^2

SS_Treatment =1209.87

SSerror - Σ (ni – 1) και si?)

SS_Error = 10764.63

SS_total = SS_Treatment + SS_Error = 1209.87 + 10764.63 = 11974.51

MS_treatment  =1209.87 / 3 = 403.29

MS_Error =10764.63 /37 = 299.0175

F =403.29 / 299.0175

F = 1.3

a) Test statistic :

F= 1.3

Source	DF	Sum of Squares	Mean square (MS)	F
Treatment	K-1 =3	SSTreatment =1209.87	MSTreatment 403.29	MSTreatment/MSError  =1.3
Error	N-k = 37	SSError =10764.63	MSError=299.0175
Total	N-1=39	SStotal = 11974.51

P value : Here significance level = 0.05 and F=1.3

p-value = 0.28893

Since P- value > 0.05 , then we Fail to reject null hypothesis at 0.05 significance level

b) using either classical approach or p value approach , you will fail to reject null hypothesis

Ans :using either classical approach or p value approach , you B