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Unformatted text preview: TTest of mean difference = 0 (vs not = 0): TValue = 1.91 PValue = 0.069 (c) H(0) : husband(mean) = wife(mean) H(a) : husband(mean) > wife(mean) With a significance level of 0.05, again we fail to reject the null hypothesis (d) but the Pvalue this time is 0.069 versus 0.647 a dramatic difference from using unpaired data. Using the Paired T we get a much more accurate portrayal of the difference and if we had used a large sample size it is likely we would have accepted the alternative hypothesis instead. (e) Husbands Age wife's age Difference in Ages Mean 35.71 33.83 1.875 SD 14.56 13.56 4.812 (f) Variable N Mean StDev SE Mean 95% CI Difference 24 1.875 4.812 0.982 (0.157, 3.907) This interval does include the value zero, therefore could not accept the alternative hypothesis that the mean male age is greater than the mean female age. Activity 4...
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 Spring '05
 staff
 Statistics, Statistical hypothesis testing, 0.982 95%, 1.87 95%, 2.97 2.77 0.982 95%

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