# Statistical Analysis HIV Testing Data and Its Implications

In this comprehensive statistical analysis homework, we delve into a dataset concerning HIV testing, treatment effects, and various statistical hypotheses. Through a series of problems, we calculate proportions, perform hypothesis tests, and even explore causal relationships. Join us on this data-driven journey as we uncover insights into HIV testing outcomes and its significant impact on healthcare decisions.

## Problem Description:

In this homework, we are analyzing a dataset to make inferences about proportions and test hypotheses. The data relates to HIV testing and the effect of a treatment on patient outcomes. We will calculate confidence intervals, perform hypothesis tests, and draw conclusions about the data.

## Solution

Problem 1

a.

 X 271 n 324 p 0.8364

This implies that at 5% level of significance, we are confident that the sample proportion lies between 0.7962 and 0.8766

Level of significance: a= 0.05

e.

Treatment Control Total
HIV Positive 13 40 53
HIV Negative 151 120 271
Total 164 160 324

g.

Computation:

h.

It is an experimental study. Treating it as a causal relationship means that one variable is believed to have effect or influence the outcome of another variable.

Problem 2

• 2 x 3 table
Age
Approval 18-34 35-54 55+ Total
Approved 68 128 173 369
Not approved 164 204 165 533
Total 232 332 338 902
• Hypothesis:
• Expected frequency
Age
Approval 18-34 35-54 55+ Total
Approved 94.9091 135.8182 138.2727 369
Not approved 137.0909 196.1818 199.7273 533
Total 232 332 338 902

df = 2, p-value = 0.0000

reject the null hypothesis and conclude that there is an association between age of the respondent and approval of the “stop and frisk” policy.

• From the analysis carried out, there exists as association between the approval of the “stop and frisk” policy and age of the respondent. It is recommended that political activist consider age while approving police stop and frisk” policy since both are dependent.

Problem 3

• Hypothesis:
• Expected count
Dates Counts Probability Expected counts 2
6th 315641 0.33333333 312340.33 34.8799
13th 298749 0.33333333 312340.33 591.4201
20th 322631 0.33333333 312340.33 339.0463
Total 937021 1 937021 965.3462

Since P-value is less than 5%, we reject the null hypothesis.

• Based on the goodness of fit carried out, we can conclude that people’s baby choices with respect to Fridays (6th, 13th, and 20th) are not equally likely.

Problem 4

When the mother smokes 0 cigarettes/day

birth weight: 119.77-0.514*0=119.77

the predicted birth weight when the mother smokes 0 cigarettes/day is 119.77

When cigarettess/day=20

birth weight= 119.77-0.514*20= 109.49

When compared to part A, we noticed a decrease in the baby’s weight from 119.77 77 when no cigarette was smoked per day to 109.49 which is a reduction of 10.28 ounces when 20 cigarettes was consumed per day.

Yes, the regression results do capture a causal relationship between average number of cigarettes smoked by mother during pregnancy and infant birth weight. As the number of cigarettes consumed increases, infant birth weight decreases.

Mother’s weight is missing from the analysis as weight of mother during pregnancy also influence or tends to increase the baby’s weight.