Gessner and Healey administered questionnaires to samples of undergraduates. Among other things, the questionnaires contained a scale that measured attitudes toward interpersonal violence (higher...

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in
Since given sample sizes are large, we can assume sample standard deviations to
be equal to population standard deviations σ. (Typically sample size is consid-
ered large when n ≥ 30)
Part A
Test: µ1 6= µ2 using two sample z-test
Hypothesis test:
H0 : µ1 − µ2 = 0 (Null Hypothesis)
Ha : µ1 − µ2 6= 0 (Alternative Hypothesis, also called H1)
This is two tailed test.
sample 1 sample 2
x1 = 2.99 x2 = 2.29
σ1 = 0.88 σ2 = 0.91
n1 = 122 n2 = 251
Significance Level, α = 0.05
Since Variances are known, we use 2 sample z-test
Standard Error: σx1−x2 =
√
σ21
n1
+ σ
2
2
n2
Test statistic, zobserved =
x1 −x2√
σ21/n1 + σ22/n2
zo =
2.99− 2.29
0.7744/122 + 0.8281/251 ≈ 7.127015 ≈ 7.13
Test statistic, zo = 7.13
Using excel we find p−value = 2∗P (Z > |7.127|) = 2*normdist(-7.127, 0,1,True
) = 0.0000000000
Critical Value, zcritical = zα/2 = z0.025 = 1.96 (From z table)
Critical value, zc = ±1.96
Rejection criteria: Rejection region is in direction of Alternative hypothesis,
since Ha : µ1 − µ2 6= 0, rejection rule is:
Reject H0 if |zo| > zc
Decision: Since |zo| > zc or equivalently since p − value < α, we reject the
null hypothesis. Hence at 5% significance level, we have sufficient evidence to
conclude that µ1 is different from µ2
Part B
Test: µ1 6= µ2 using...