Question 1

Question 1 1(a) Describe the structure of the study (independent variables, dependent variable) and detail why the researchers would have used random allocation. 1(b) Produce a table providing summary statistics to compare the four groups. Produce at least one graphic to compare the four groups. Comment on what your table and graphic show. Descriptive Statistics Dependent Variable: Calcium levels Mum Treatment Mean Std. Deviation N Breast Supplement 2.6398 .33698 82 Placebo 2.3737 .33914 82 Total 2.5067 .36249 164 Bottle Supplement 2.2178 .36224 82 Placebo 2.1446 .36171 82 Total 2.1812 .36273 164 Total Supplement 2.4288 .40795 164 Placebo 2.2591 .36792 164 Total 2.3440 .39705 328 Table 1. Figure 1. Boxplot chart Figure 3. Error Bars 1(C) Perform a two-way analysis on the data provided. What additional conclusions can be drawn from the two-way analysis? Table 1, Descriptive Statistics Dependent Variable: Calcium levels Mum Treatment Mean Std. Deviation N Breast Supplement 2.6398 .33698 82 Placebo 2.3737 .33914 82 Total 2.5067 .36249 164 Bottle Supplement 2.2178 .36224 82 Placebo 2.1446 .36171 82 Total 2.1812 .36273 164 Total Supplement 2.4288 .40795 164 Placebo 2.2591 .36792 164 Total 2.3440 .39705 328 Table 2, Tests of Between-Subjects Effects Dependent Variable: Calcium Source Type III Sum of Squares df Mean Square F Sig. Mum 8.687 1 8.687 70.826 .000 Treatment 2.360 1 2.360 19.237 .000 Mum * Treatment .763 1 .763 6.221 .013 Error 39.741 324 .123 a. R Squared = .229 (Adjusted R Squared = .222) Table 3, Estimated marginal mean for feeding methods and treatment A. Mum Dependent Variable: Calcium levels Mum Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound Breast 2.507 .027 2.453 2.561 Bottle 2.181 .027 2.127 2.235 B. Treatment Dependent Variable: Calcium levels Treatment Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound Supplement 2.429 .027 2.375 2.483 Placebo 2.259 .027 2.205 2.313 Table 3 (A,B). Figure 1. 1(d) Group Statistics group N Mean Std. Deviation Std. Error Mean Calcium breast and supplement 82 2.6398 .33698 .03721 breast and placebo 80 2.2295 .35828 .04006 Table 1. Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df p Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper Calcium Equal variances assumed .221 .639 7.509 160 .000 .41026 .05463 .30236 .51815 Equal variances not assumed 7.503 158.825 .000 .41026 .05468 .30227 .51824 Question 2 2(a) Please paraphrasing the answer from the paper, which I upload it already 2(b) Smoking Status * Disease Status Crosstabulation Smoking Status Total Non-smoker Smoker Disease Status Resolved 211 95 306 Persisted 81 57 138 Treated 68 54 122 Total 360 206 566 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 8.481a 2 .014 Likelihood Ratio 8.464 2 .015 Linear-by-Linear Association 7.898 1 .005 N of Valid Cases 566 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 44.40. Bar Chart 2(c) Table 1, Disease Outcome * Smear Test Crosstabulation Smear Test Total No dyskaryosis Dyskaryosis Disease Outcome Persisted 125 135 260 Resolved 238 68 306 Total 363 203 566 Table 2, Chi-Square Tests Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square 53.907a 1 .000 Continuity Correctionb 52.624 1 .000 Likelihood Ratio 54.558 1 .000 Fisher's Exact Test .000 .000 Linear-by-Linear Association 53.812 1 .000 N of Valid Cases 566 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 93.25. b. Computed only for a 2x2 table Odds ratio relative risk (Cohranss & Mantel-haenszel statistics) Tests of Homogeneity of the Odds Ratio Chi-Squared df Asymp. Sig. (2-sided) Breslow-Day .000 0 . Tarone's .000 0 . Tests of Conditional Independence Chi-Squared df Asymp. Sig. (2-sided) Cochran's 53.907 1 .000 Mantel-Haenszel 52.531 1 .000 Under the conditional independence assumption, Cochran's statistic is asymptotically distributed as a 1 df chi-squared distribution, only if the number of strata is fixed, while the Mantel-Haenszel statistic is always asymptotically distributed as a 1 df chi-squared distribution. Note that the continuity correction is removed from the Mantel-Haenszel statistic when the sum of the differences between the observed and the expected is 0. Mantel-Haenszel Common Odds Ratio Estimate Estimate .265 ln(Estimate) -1.330 Std. Error of ln(Estimate) .185 Asymp. Sig. (2-sided) .000 Asymp. 95% Confidence Interval Common Odds Ratio Lower Bound .184 Upper Bound .380 ln(Common Odds Ratio) Lower Bound -1.693 Upper Bound -.967 The Mantel-Haenszel common odds ratio estimate is asymptotically normally distributed under the common odds ratio of 1.000 assumption. So is the natural log of the estimate. 2(d) Please paraphrasing the answer from the paper, which I already upload it but please use the number from my results.