Question

1- A1C level is related to blood sugar and is measured for patients who have Type 2 diabetes or a tendency to have it. It should normally bet

A1C Before	A1C After
7.9	6.9
7.8	7.1
8.3	7.5
7.4	7.3
7.3	7.3
7.4	7.5
7.8	6.8
7.6	6.7
7.3	6.8
7.4	6.7
7.3	7.1
7.4	7.0
7.2	7.4

Test, at level 0.05, whether the diet/exercise regimen has lowered the A1C level of a patient.

a) The null hypothesis is

b) The value of the test statistic is

c)  The p-value of the test is

2- In a study, 566 patients were treated with a medication for a disease of which 202 recovered fully. In the same study, 365 patients were given placebo (an inert substance that looks like the medication), 96 recovered fully.

a) Obtain a 95% confidence interval for the difference between proportions of patients who recovered fully for both groups.

b) Test whether there is difference between proportions of patients who recovered fully for both groups. Use α=0.05

Answer 1

1.

(a) Null Hypothesis: The AIC level before and after the diet / exercise regimen are the same

Alternative Hypothesis: The AIC level before diet/ exercise regimen is more.

As the same units are treated before and after we are applying paired t test

t-Test: Paired Two Sample for Means
	Variable 1	Variable 2
Mean	7.546154	7.084615
Variance	0.101026	0.08641
Observations	13	13
Pearson Correlation	0.03499
Hypothesized Mean Difference	0
df	12
t Stat	3.912581
P(T<=t) one-tail	0.001031
t Critical one-tail	1.782288
P(T<=t) two-tail	0.002063
t Critical two-tail	2.178813

(b) The value of t statistic is 3.912581

(c) It is a one tailed test so p value is .001031

Since we are testing at 5% level of significance and the p value is .001031. So the null hypothesis is rejected and we can conclude that the diet/ exercise regimen has lowered the AIC level

2.

Groups	I	II
Number	566	365
Recovered Fully	202	96

p1= 202/566 and p2= 96/365

The combined proportion p= (n1p1+ n2p2)/ (n1+n2)= (202/566*566+96/365*365)/ (566+365)=0,32

q=(1-p)= 1-0.32=0.68

Standard Error (S.E)= sqrt{p*q*(1/n1 + 1/n2)}= 0.0313

(a) the confidence interval (p1-p2)(+/- )1.96 (S.E of p1-p2)

1.96 is z value for 5% level of signifiacance for a two tailed test

The upper confidence limit is 0.1552 and the lower confidence limit is 0.0939 of fully recovered patients of two groups

(b) Ho: The proportions of fully recovered patients are equal in two groups

Ha: The proportions of fully recovered patients are not equal in two groups

This is a two tailed test

IzI= (p1-p2)/ S.E= (202/566-96/365)/.0313= 2.9992 with p value .0026

As we are testing at 5% level of significance null hypothesis is rejected since p=.0026 is less than 0.05. So we can conclude that the proportions of fully recovered patients are not equal in two groups