Coursework Assessment specification

Order Instructions/Description

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
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.