Question 1

| December 1, 2015

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 2×2 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.

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