Question

The statistics lecture room is divided into three sections: front, middle, and back. The instructor noticed that the further the students were from her, the more likely they were to miss class, arrive late or text during class. She wanted to see if the students who sit further away did worse on tests so she took a random sample of students from each section and recorded their percentage mark on the second test. Percentage Mark on Second Test Front Middle Back (6 numbers) (8 numbers) (7 numbers) 38 78 59 97 68 60 93 81 83 83 85 59 78 45 77 67 51 75 70 73 | Average 80 58 (a) Calculate the grand (overall) mean. [2 marks] Grand Mean (b) Calculate SSB. [3 marks] SSB (c) Just calculate the contribution to SSW from the 'Front' group and use the results given for the other two to calculate the value of SSW. [3 marks] | Front x - X Front Front Middle Back x 81 83 97 93 59 67 Total 696 1,312 SSW =

Answer 1

Ans.

Front 38 middle back 81 83 83 78 68 851 59 60 971 zo Average SS 80 1078 1680 7358 696 1312 70 3086 SSB

a. Grand Mean = 70

b. Using the given formulae we find the value of SSB= 1680

c. SSB = 3086

We are given the sum of square for middle and Back and by using Excel (using SSF formulae we find the value of ssf).

the sum of SSF+SSM+SSB = SSW.

d.  

p.val 0 Source of Variation SS Between Groups Within Groups Total MS Ftest 2 840 4.899546 18 171.4444 Fcrit .02 3.555 D F 1680 3086 4766 20

We can see that our F-val > F crit also P-value <0.05 , which implies we will reject our null hypothesis i.e. there is sufficient evidence to say that the variation Between Groups is significant .