Question

A company is trying to decide which of two types of tires to buy for their trucks.

Test whether the differences in scores may be attributed to chance , at 0.01 significance level. the mathematical scores of 6 groups of 4 students each are shown below.

G1 G2 G3 G4 G5 G6

Student 1 84 90 70 95 85 93

Student 2 88 95 73 96 78 85

Student 3 70 93 85 90 75 80

Student 4 80 80 90 90 90 90

Answer 1

1. Means in groups

$m_1=\frac{84+88+70+80}{4}=80.5$

$m_2=\frac{90+95+93+80}{4}=89.5$

$m_3=\frac{70+73+85+90}{4}=79.5$

$m_4=\frac{95+96+90+90}{4}=92.75$

$m_5=\frac{85+78+75+90}{4}=82$

$m_6=\frac{93+85+80+90}{4}=87$

2 General mean

$m=\frac{80.5+89.5+79.5+92.75+82+87}{6}=85.207$

3 Sum of squares between groups

$SSB=4\cdot ( (80.50-85.207)^2 + (89.50-85.207)^2+\\ (79.50-85.207)^2 + \\ (92.75-85.207)^2+ (82-85.207)^2+(87-85.207)^2 )=574.208$

4 Sum of squares in groups

$SSR_1=(84-80.5)^2+(88-80.5)^2+(70-80.5)^2+\\(80-80.5)^2=179$

$SSR_2=(90-89.5)^2+(95-89.5)^2+(93-89.5)^2+\\(80-89.5)^2=133$

$SSR_3=(70-79.5)^2+(73-79.5)^2+(85-79.5)^2+\\(90-79.5)^2=273$

$SSR_4=(95-92.75)^2+(96-92.75)^2+(90-92.75)^2+\\(90-92.75)^2=30.75$

$SSR_5=(85-82)^2+(78-82)^2+(75-82)^2+\\(90-82)^2=138$

$SSR_6=(93-87)^2+(85-87)^2+(80-87)^2+\\(90-87)^2=98$

SSR=179+133+273+30.75+138+98=851.75

5 F-criterion

$\frac{SSB/(k-1)}{SSR/(n-k)}=\frac{\frac{574.208}{5}}{\frac{881.75}{18}}=2.131$

Table value of F criterion

Ftab=[soft Mathcad]=qF(1-0.01,5,18)=4.248

6 Conclusion

F<Ftab therefore there are no sufficient grounds to establish differences of scores between 6 groups at significance level 0.01