Question

Colonoscopy is a medical procedure that is designed to find and remove precancerous lesions in the colon before they become cancerous. In a sample of 51,460 people without colorectal cancer, 5043 had previously had a colonoscopy and in a sample of 10,292 people diagnosed with colorectal cancer, 720 had previously had a colonoscopy. A test of the claim that the proportion of people who have had colonoscopies is the same among those with colorectal cancer and those without will be performed, using significance level 0.01, by answering the questions below.

a) What are the observational units?

b) What is the response variable here?

c) Describe the group variable.

d) Find the proportion within each group that had a colonoscopy.

e) Find the overall proportion of individuals across both groups that had a colonoscopy.

f) What are the appropriate null and alternative hypothesis for testing the claim?

g) Calculate the test statistics, showing your work.

h) What is the p-value for the test?

i) What is your decision based on the p-value? Please be clear in stating the decision rule and making the comparison between the p-value and the significance level.

j) State your conclusion in the context of the problem.

Answer 1

Here we have given that,

(A) observational units

Claim: To check whether the difference in the proportion of people who have had colonoscopies is the same among those with colorectal cancer and those without will be performed .

(B) response variables

P1=proportion of people who have had colonoscopies with colorectal cancer

P2=proportion of people who have had colonoscopies without colorectal cancer

(C) group variables

we have given that,

n1= number of people with colorectal cancer =51460

x1=number of people previously had colonoscopy=5043

n2=number of people diagonsed with colorectal cancer=10292

X2=number of people previously had colonoscopy=720

= level of significance= 0.01

(D) Finding proportion within each group

Now we estimate the proportion p as

=1^st sample proportion =

= 2^nd sample proportion =

(e) overall proportion across both groups

=combine proportion of two samples == =0.093

(f) Null and alternative hypothesis

The Hypotheses is

Ho: P1=P2

v/s

H1:P1 P2

Now, we can find the test statistics

= 9.89

Now,

Pvalue = 2*P( Z> 9.89) This is two tailed test

              = 2* (1- P (Z < 9.89))

              = 2 * ( 1- 1.000))

              =2*0

              =0.000

Decision:

Here Pvalue< 0.01

Here we can reject the Null hypothesis Ho

Conclusion:

That is here there is sufficient evidence that there is difference in the proportion of people who have had colonoscopies is the same among those with colorectal cancer and those without will be performed