This study is conducted in order to know the effectiveness of a new anti-anxiety

Question

This study is conducted in order to know the effectiveness of a new anti-anxiety medication that whether the male and female patients are equally are recovered or not. Data: The data is assumed for this study from a web source (Free Statistics Lectures). Data: Out of 200 respondents 64 male patients are asked about the cure of new anti-anxiety medication and its effectiveness and similarly, out of 200 respondents, 92 female patients are asked about the effectiveness of the cure of new anti-anxiety medication as well. Therefore, the male patients are denoted by p1 and female patients are denoted by p2 Calculate and find P-Value and 95% Confidence Interval.

Explanation / Answer

(a)

Data:   

n1 = 200

n2 = 200

p1 = 64/200 = 0.32

p2 = 92/200 = 0.46

Hypotheses:   

Ho: p1 = p2   

Ha: p1 p2   

Decision Rule:   

= 0.05

Lower Critical z- score =   -1.959963985

Upper Critical z- score = 1.959963985

Reject Ho if |z| >   1.959963985

Test Statistic:   

Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (200 * 0.32 + 200 * 0.46)/(200 + 200) = 0.39

q = 1 - p = 1 - 0.39 = 0.61

SE = [pq * {(1/n1) + (1/n2)}] = (0.39 * 0.61 * ((1/200) + (1/200))) = 0.048774994

z = (p1 - p2)/SE = (0.32 - 0.46)/0.0487749935930288 = -2.870323288

p- value = 0.004100523

(b)

n1 = 200

n2 = 200

p1 = 0.32

p2 = 0.46

% = 90

Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (200 * 0.32 + 200 * 0.46)/(200 + 200) = 0.39

q = 1 - p = 1 - 0.39 = 0.61

SE = (pq * ((1/n1) + (1/n2))) = (0.39 * 0.61 * ((1/200) + (1/200))) = 0.048774994

z- score = 1.644853627

Width of the confidence interval = z * SE = 1.64485362695147 * 0.0487749935930288 = 0.080227725

Lower Limit of the confidence interval = (p1 - p2) - width = -0.14 - 0.0802277251160283 = -0.220227725

Upper Limit of the confidence interval = (p1 - p2) + width = -0.14 + 0.0802277251160283 = -0.059772275

The 90% confidence interval is [-0.22, -0.06]

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