# 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 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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