Frequency distribution, cross-tabulation, and hypothesis testing

  1. Open file
  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

  1. If the table is 2x2 or nij <=10  Continuity correction needed
  2. Phi (2x2)
  3. Contingency coefficient C (rxc). where r or c >2
  4. Cramer’s V (rxc), where r or c >2
  1. 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 ≠ 4.0Then, p-value = sig. (the value "sig." is found in the the SPSS table)

b) H1: μ > 4.0Then, p-value = sig./2

c) H1: μ < 4.0Then, p-value = 1 - (sig./2)

II) When the value of statistic t in the SPSS computer printout is less than zero (t < 0)

andyour alternative hypothesis H1

a) H1: μ ≠ 4.0Then, p-value = sig. (the same as in Scenario I)

b) H1: μ > 4.0Then, p-value = 1 - (sig./2)

c) H1: μ < 4.0Then, p-value = sig./2.

  1. Two independent samples: AnalyzeCompare MeansIndependent-Samples t- Test
  2. Test variable: INT_USAGE
  3. Grouping variable: GENDERDefine Groups: Group 1: 1; Group 2: 2
  4. 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)

  1. Note: because F test rejects the aboveH0a, 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”).

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.

  1. 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)

  1. Kolmogorov-Smirnov (K-S): One sample test whether a distribution is normal.
  2. Exercise: test the distribution of INT_FAMILIAR

H0: The distribution of INT_FAMILIAR is normal

  1. AnalyzeNonparametric Tests1-Sample K-S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 H0when, 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 makea mistake and reject a good H0.

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