Question

The University of Ottawa is interested in offering its employees one of two employee
benefit packages. A random sample of the university’s employees is collected, and each
person in the sample is asked to rate each of the two packages on an overall preference
scale of 0 to 100. The results are
package A: 45, 67, 63, 50, 77, 60, 47, 39, 56, 68, 70
package B: 56, 79, 60, 45, 85, 39, 50, 41, 50, 69, 82
After analyzing the data, the University concludes that its employees prefer, on
average, one package over the other, i.e., there is a significant difference between the
two packages, at significance level = 0:05.
True False

The University of Ottawa is interested in offering its employees one of two employee benefit packages. A random sample of the university's employees is collected, and each person in the sample is asked to rate each of the two packages on an overall preference scale of 0 to 100. The results are package A: package B: 45, 67, 63, 50, 77, 60, 47, 39, 56, 68, 70 56, 79, 60, 45, 85, 39, 50, 41, 50, 69, 82 After analyzing the data, the University concludes that its employees prefer, on average, one package over the other, i.e., there is a significant difference between the two packages, at significance level a 0.05. = True False

Answer 1

Solution :  False

Explanation :

Let ,The random sample of Package A and Package B

Hypothesis :

Null Hypothesis : Mean difference of two packages are equal.

Alternative Hypothesis : Mean difference of two packages are not equal.

Mathematically :

Ho: M1 = 42

211111: "H

Where
11
and
112
are the mean of package A and package B respectively.

Test Statistic:

Here we have sample size is 11 and Population standard deviation not know so we use "T - test"

(X1-X) t: +影) 21

Where S = Pooled standard deviation

n1 , n2 are sample sizes and

X1
and
X
are sample means

S = (n1 - 1) * S + (n2 - 1) * S22 ni + nj-2

s = 14.5346

Test statistic becomes;

(58.4-59.6) t: 14.5346 * *v( + )

t=-0.21

t = 0.21

Critical T :

tcrit tn +n2-1,a/2 = t20,0.025

From T table :

t20,0.025 2.086

Criteria : If |t| < t_crit , then accept the null hypothesis at level of significance.

So here 0.21 < 2.086 , we may accept the null hypothesis at 5% level of significance.

Conclusion :

The average of package A does not differ to average of package B

So there is no significant difference between two packages.

Minitab Output (for reference) :

Two-Sample T-Test and CI: A, B

Two-sample T for A vs B

N Mean StDev SE Mean
A 11 58.4 12.0 3.6
B 11 59.6 16.7 5.0

Difference = μ (A) - μ (B)
Estimate for difference: -1.27
95% CI for difference: (-14.20, 11.66)
T-Test of difference = 0 (vs ≠): T-Value = -0.21 P-Value = 0.839 DF = 20
Both use Pooled StDev = 14.5346