Strong Relationship between Two Data Sets - Essay Example

Paired T for Sample 2 - Sample 3N Mean StDev SE MeanSample 2 19 1.5747 0.3893 0.0893Sample 3 19 1.5927 0.3686 0.0846Difference 19 -0.0181 0.2493 0.057295% CI for mean difference: (-0.1382, 0.1021)T-Test of mean difference = 0 (vs not = 0): T-Value = -0.32 P-Value = 0.756The paired-t-test has been selected because we needed to test whether the two samples had different means and if these differences were significant. We, therefore, needed to pair these two samples and test the differences at 95% confidence level.

As can be seen, p > .05 therefore the differences are not significant. Caffeine Consumption and Marital Status0 1-150 151-300 >300 TotalMarried 652 1537 598 242 3029705.83 1488.01 578.07 257.094.106 1.613 0.687 0.886Divorced 36 46 38 21 14132.86 69.27 26.91 11.970.301 7.815 4.571 6.817Single 218 327 106 67 718167.31 352.72 137.03 60.9415.356 1.876 7.025 0.602Total 906 1910 742 330 3888Chi-Sq = 51.656, DF = 6, P-Value = 0.000The chi-square analysis was run for the categorical values as were in the table.

The chi-square tested the hypothesis that there is no association between caffeine consumption (column factors) and marital status (raw factors). The results show that the chi-square statistic was 51.656. The p-value is < .001 which suggests that we reject the null hypothesis of no association between marital status and caffeine consumption in favor of the alternative that there is indeed an association between them. Thus, marital status affects caffeine consumption. The reason for this association could be attributed to the fact that we had more married participants in the sample and hence the results may be biased towards the same.

As can be seen, there were 3029 married participants out of the 3888 participants in total. I recommend that to improve results, the same number of samples should be selected from the marital status categories for the study.