Question

1. The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 411 of 1012 mean and 535 of 1062 women suffered from some form of arthritis.

1. Test the claim that a different proportion of senior men and women who have this disease at the α = 0.05.

Answer 1

Answer)

Null hypothesis Ho : p1 = p2

Alternate hypothesis Ha : p1 not equal to p2

N1 = 1012 , P1 = 411/1012

N2 = 1062, P2 = 535/1062

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 411

N1*(1-p1) = 601

N2*p2 = 535

N2*(1-p2) = 527

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = -4.46

From z table, P(Z<-4.46) = 0

Since the test is two tailed

So, P-Value = 2*0 = 0

As the obtained P-Value is less than the given significance level

We reject the null hypothesis Ho

So, we have enough evidence to conclude that difference exist