Download HypTest

Frequency distribution, cross-tabulation, and hypothesis testing
1. Open file http://spartan.ac.brocku.ca/~ekaciak/ABR1/InternetHypTest.xls
2. Variable ViewVariable INT_FAMILIARMissing9
3. AnalyzeDescriptive StatisticsFrequenciesINT_FAMILIARcheck all
boxes
4. TransformRecodeInto Different variables(INT_FAMILIAR -> INTnew)
Old and New ValuesRange 1 through 5 -->1 and Range 6 through 7-->2
5. AnalyzeDescriptive StatisticsCrosstabsGENDERColumns(s) – because
GENDER is independent; INTnewRows – because INTnew is
dependent(s)Statistics (check Nominal only)
6. Test with X2 (only on counts data, not %, good for nij>=5)
H0: There is no association between the two variables
a. If the table is 2x2 or nij <=10  Continuity correction needed
b. Phi (2x2)
c. Contingency coefficient C (rxc). where r or c >2
d. Cramer’s V (rxc), where r or c >2
7. AnalyzeCompare MeansOne-Sample t-testINT_FAMILIARCI=95%,
Test Value = 4.0
Note: There are two possible scenarios (in each of them, H0: μ = 4.0)
I) When the value of statistic t in the SPSS computer printout is greater than zero (t > 0)
and your alternative hypothesis H1
a) H1: μ ≠ 4.0 Then, p-value = sig. (the value "sig." is found in the the SPSS table)
b) H1: μ > 4.0 Then, p-value = sig./2
c) H1: μ < 4.0 Then, p-value = 1 - (sig./2)
II) When the value of statistic t in the SPSS computer printout is less than zero (t < 0)
and your alternative hypothesis H1
a) H1: μ ≠ 4.0 Then, p-value = sig. (the same as in Scenario I)
b) H1: μ > 4.0 Then, p-value = 1 - (sig./2)
c) H1: μ < 4.0 Then, p-value = sig./2.
8. Two independent samples: AnalyzeCompare MeansIndependent-Samples tTest
a. Test variable: INT_USAGE
b. Grouping variable: GENDERDefine Groups: Group 1: 1; Group 2: 2
i. Note: F test rejects H0a: The two populations have equal variances,
at the significance level of 0.05.
Why? From SPSS, Sig. = 0.000, which is less than alpha =
0.05 reject H0a (the two variances are statistically different at
the significance level of 0.05 or less)
ii. Note: because F test rejects the above H0a, in order to test our main
hypothesis H0: The two populations have equal means, at the
significance level of 0.05, we use the case with “Equal variances
not assumed”).
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Again, in order to make a decision whether to reject H0 or not,
we follow the procedure: From SPSS, Sig. 0.000, which is less
than alpha = 0.05  reject H0. The means are statistically
different at the significance level of 0.05 or less.
9. Paired samples: AnalyzeCompare MeansPaired Samples t-test
INT_ATTITUDE vs. TECHN_ATTITUDE
Sig. = 0.000, thus reject H0: The difference between the two means is zero at
the confidence level of 0.000 (which is well below the typical 0.05)
10. Kolmogorov-Smirnov (K-S): One sample test whether a distribution is normal.
a. Exercise: test the distribution of INT_FAMILIAR
H0: The distribution of INT_FAMILIAR is normal
i. AnalyzeNonparametric Tests1-Sample KSINT_FAMILIARcheck the box NormalStatistics:
DescriptivesOK
Since Sig. is 0.178, we cannot reject H0 at the significance
level of 0.05. In order to reject H0, Sig. would have to be
less than the significance level of 0.05.
Attention: If you cannot reject H0, never say: “Therefore, we
accept H0”, although it sounds similar. Why? In this and all
other statistical tests, we can control only for Type I Error =
Reject H0 when, in fact, H0 is CORRECT. We do not control
for Type II Error = Accept H0 when, in fact, H0 is WRONG.
Remember also that the Probability of Type I Error = The
Significance Level alpha, e.g. 0.05, which means that when
rejecting H0 100 times based on 100 samples from this
population, 5 times we will make a mistake and reject a good
H0.
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