Question

9) (CH. 9-2) Is there a difference between the average NBA Championship Final game winning scores of the 1970’s versus the average of the winning scores of the 2000’s? Use a 0.01 significance level to test the claim that there is a difference.

	1970’s	2000’s
	97	99
	105	131
	109	83
	87	95
	96	81
	102	100
	102	88
	114	113
	118	108
	113	116

Use the data from problem #9 to construct a 99% confidence interval estimate for the mean of the differences. Does this interval contain zero? Do the results of this problem support the results of problem #9?

• What is the Confidence Interval Estimate:

• Does the interval indicate a difference or not? Explain your answer.

• Do the results of the hypothesis test (from problem #9) and the confidence interval estimate support each other?

Answer 1

From the data:

	1970's	2000's
n	10	10
Sum	1043	1014
Average	104.30	101.4
SS(Sum of squares)	812.1	2250.4
Variance = SS/n-1	90.23	250.04
Std Dev=Sqrt(Variance)	9.5	15.81

Since s₁/s₂ = 0.6 (it lies between 0.5 and 2) we used the pooled variance

The Pooled Variance is given by:

Sp = 170.1031

= 0.01

The Hypothesis:

H0:

Ha:

This is a Two tailed test.

The Test Statistic:We use the students t test as population standard deviations are unknown.

The p Value:    The p value (2 Tail) for t = 0.497, df = 18, is; p value = 0.6252

The Critical Value:   The critical value (2 tail) at = 0.01,df = 18, tcritical= +2.878 and -2.878

The Decision Rule: If tobservedis >tcriticalor If tobserved is < -tcritical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision: Since t lies in between +2.878 and -2.878, We Fail To Reject H0

Also since P value (0.6252) is > (0.01), We Fail to Reject H0.

The Conclusion: There is insufficient evidence at the 99% significance level to conclude that there is a significant difference in the means of the NBA championship final scores from 1970's and the 2000's.

_____________________________________________________________________

The 99% CI

The tcritical (2 tail) for = 0.01, df = n₁ + n₂ – 2 = 10 + 10 – 2 = 18 is 2.878

The Confidence Interval is given by (- ) ME, where

(- ) = 104.3 – 101.4 = 2.9

The Lower Limit = 2.9 - 16.787 = -13.887

The Upper Limit = 2.9 + 16.787 = 19.687

The 99% Confidence Interval is (-13.887 , 19.687)

The interval contains 0. Therefore there is a possibility for the means to be equal and hence, we fail to reject H0. Hence there is no significant difference.

Yes, the results from the hypothesis test and the confidence interval support each other.