Question

answer Question 3 (16)

• (a) ANOVA output: 3 points

• (b) Null and alternative hypotheses in terms of the population means: 3 points Pooled estimate of the common variance: 2 points

• Value of the test statistic: 2 points

• Null distribution: 2 points

• The P-value: 2 points

• Conclusion in plain language: 2 points

the following is the data use stat crunch or excel

Category code score 
C 1 45
C 1 53
C 1 72
C 1 55
C 1 67
C 1 53
C 1 66
C 1 54
C 1 53
C 1 67
C 1 60
C 1 64
C 1 69
C 1 52
C 1 74
C 1 58
C 1 53
C 1 66
C 1 66
C 1 71
H 2 79
H 2 95
H 2 81
H 2 74
H 2 68
H 2 84
H 2 78
H 2 77
H 2 85
H 2 72
H 2 86
H 2 71
H 2 67
H 2 77
H 2 83
H 2 73
H 2 76
H 2 86
H 2 83
H 2 75
Q 3 58
Q 3 66
Q 3 67
Q 3 68
Q 3 64
Q 3 57
Q 3 58
Q 3 57
Q 3 75
Q 3 60
Q 3 63
Q 3 55
Q 3 64
Q 3 51
Q 3 60
Q 3 55
Q 3 61
Q 3 63
Q 3 76
Q 3 64
T 4 51
T 4 57
T 4 73
T 4 67
T 4 63
T 4 77
T 4 67
T 4 81
T 4 70
T 4 82
T 4 61
T 4 58
T 4 80
T 4 68
T 4 76
T 4 60
T 4 69
T 4 68
T 4 86
T 4 64
HT 5 78
HT 5 85
HT 5 86
HT 5 92
HT 5 84
HT 5 78
HT 5 79
HT 5 78
HT 5 95
HT 5 81
HT 5 83
HT 5 76
HT 5 83
HT 5 72
HT 5 81
HT 5 70
HT 5 81
HT 5 90
HT 5 95
HT 5 84
QT 6 55
QT 6 80
QT 6 67
QT 6 89
QT 6 90
QT 6 72
QT 6 75
QT 6 61
QT 6 68
QT 6 56
QT 6 63
QT 6 66
QT 6 64
QT 6 65
QT 6 62
QT 6 70
QT 6 83
QT 6 72
QT 6 65
QT 6 73

Answer 1

(a)

79 51 Anova: Single Factor 53 72 57 81 SUMMARY 92 Grou Count Sum Average Variance 1218 1570 1242 1378 1651 1396 67 53 60.9 68.30526316 78.5 48.36842105 62.1 40.30526316 68.9 87.25263158 82.55 45.83947368 94.8 57 72 75 61 67 53 67 75 81 69.8 61 83 63 71 83 72 81 69 ANOVA 51 Source of Variation Between Groups Within Groups MS 1499.848333 23.38209106 4.16E-16 2.293911 64.14517544 P-value F crit 62 7499.241667 7312.55 73 114 53 61 81 83 Total 14811.79167 119 95 73

(b) μί-population mean of ith code, i-1.2.3.4.5.6 Null hypothesis, Ho : μ1-12-13-14-15-16 Alternative hypothesis. H1 : at least one μ¡ is different from others. Pooled estimate of the common variance - Mean sum of square due to errorMSE 64.1452 Value of the test statistic-F-233821 Null distribution: F5,114 p - value 4.1601 x 10 Conclusion: Wereject Ho since p-value too small < 0.01) hence at least one 16 is different from others.