Question

The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found 411 of 1012 men and 535 of 1062 women suffered from some sort of arthritis. Find a 90% confidence interval for the difference in the proportion of senior men and senior women who have this disease.

13. 13) Answer the following:

    1. a) Is this a “proportions of success” or a “means” problem?

    2. b) What calculator function are you using?

    3. c) What interval did you get?

Answer 1

Here, we have given that,

n1=Number of men = 1012

x1= Number of men found that they suffered from some sort of arthritis= 411

= sample proportion of men found that they suffered from some sort of arthritis

=

n2=Number of women = 1062

x2= Number of women found that they suffered from some sort of arthritis= 535

= sample proportion of women found that they suffered from some sort of arthritis

=

(A)

This problem is based on the proportions of mean and women who suffered form some sort of arthritis.

(B)

To find the critical value we will use the Excel software =NORMSINV function.

(C)

Now, we want to find the 90% confidence interval for the difference in proportion of senior men and senior women who have this disease (p1- p2).

Formula is as follows,

Now we find the Z critical value

c=confidence level = 0.90

= level of significance=1-c =1-0.90= 0.10

=

                            =  

                            = 1.645 Using EXCEL software = ABS(NORMSINV(probablity = 0.05))

The critical value is 1.645.

Now, The 90% confidence interval is

Interpretation:

This confidence interval shows that we are 90% confident that the difference in proportion of senior men and senior women who have this disease (p1- p2). is lies within that interval.