Adm2304 Mt

ADM 2304 Final Examination

Winter 2012

Question 1. [8 marks]

(a) [3] Ho: µB - µA = µd = 0; Ha: µd > 0; t = 4.0/(6.0/√9) = 2.0,

Reject Ho if t > t(.05,8) = 1.860

reject Ho. There is sufficient evidence to conclude that Food

B gives greater weight gain than Food A.

1

mark for hypothesis, 1 mark for test statistic, ½ mark for critical value or p-value, ½ for

conclusion

[if µA - µB = µd , then Ha: µd < 0 and t = -2. Then reject Ho if t < -t(.05,8) = -1.860 ]

(b) [4] Ho: µB – µA = 0; Ha: µB – µA > 0; sp = 8.7; t = (50 – 46)/ 8.7√(1/9+1/9) = 4.0/4.1 = 0.98

Reject Ho if t > t(.05,16) = 1.746 accept Ho. There is not enough evidence to conclude that Food B

gives greater weight gain than Food A.

1 mark for hypotheses, 1 mark for test statistic, 1 mark for critical value or p-value, 1 mark for

conclusion

(c)

[1] In the second test the variation between dogs masks the variation between dog foods.

Question 2 (11 marks)

a. H0: The distribution of the types of TVs sold is the same in the various provinces

Ha: The distribution of the types of TVs sold is not the same in the various provinces

The hypotheses below are acceptable, but keep in mind that we have four samples not one sample.

H0: The types of TVs sold and the provincial location are independent variables

[1]

Ha: The types of TVs sold and the provincial location are not independent variables

f.2 = 175, f2. = 214, n = 495, df = (r – 1)*(c-1) = (4-1)*(4-1) =9

E22= (f2. x f.2)/n = (214 x 175)/ 495 = 75.6566

[3]

χ222 = (O22 – E22)2/E22 = (83 – 75.6566)2/75.6566 = 0.7128

[1]

The other calculations are done similarly, and χ2Calc = 18.998

[1]

(1 mark for H0 and Ha, 2 marks for final result—1 mark for some intermediate calculations if final result

is incorrect).

b. Χ2Crit = Χ2 0.05(df = 9) = 16.919

[3]

Since { χ2Calc = 18.998} > { Χ2Crit = 16.919}

[1]

Reject H0

[1.5]

Based on the statistical evidence, the distribution of the types of TVs sold is not the same in the...