Z-TESTS We will use the formula for Z from lecture notes, page 2. since H 0 is u
ID: 3153627 • Letter: Z
Question
Z-TESTS
We will use the formula for Z from lecture notes, page 2.
since H0 is usually m1=m2, then m1-m2= 0 and:
for 1, 2 known
for 1, 2 unknown
Z = if n1 + n2 32
Steps to be covered:
State the hypotheses, and identify the claim
Find the critical value(s) – you might want to draw the curve
Compute the test (statistic) value
Make the decision to reject or not reject the null hypothesis.
PROBLEM 2:
Does smoking makes a difference when it comes to the number of absent days?
Test at .05 significance level.
Smokers: Average number of days absent = 14.7; standard deviation = 5.0; n = 44
Non-Smokers: Average number of days absent = 8.3; standard deviation = 4.0; n = 60
PROBLEM 3:
Are the machine tools manufactured by Company X and Y different with regard to how long they last?
Company X
Company Y
16.2 weeks
15.9 weeks
s
.2 weeks
.2 weeks
n
80
80
Test at a = .08
Company X
Company Y
16.2 weeks
15.9 weeks
s
.2 weeks
.2 weeks
n
80
80
Explanation / Answer
2.
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0 [claim]
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At level of significance = 0.05
As we can see, this is a two tailed test.
Hence, the critical values are
zcrit = -1.96, 1.96 [ANSWER]
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Calculating the means of each group,
X1 = 14.7
X2 = 8.3
Calculating the standard deviations of each group,
s1 = 5
s2 = 4
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 44
n2 = sample size of group 2 = 60
Also, sD = 0.913700435
Thus, the z statistic will be
z = [X1 - X2 - uD]/sD = 7.004483915 [ANSWER, TEST STATISTIC]
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where uD = hypothesized difference = 0
As |z| > 1.96, WE REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that smoking makes a difference when it comes to the number of absent days at 0.05 level. [CONCLUSION]
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