Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening program for the early detection of breast cancer was started in order to increase the survival rate p of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were screened by the program and who were diagnosed to have the disease. Let x represent the number of those in the sample who survive the disease.

Required:
a. If you wish to determine the community screening program has been effective, state the alternative hypothesis that should be tested.
b. State the null hypothesis.
c. If 164 women in the sample of 200 survive the disease, can you conclude that the community screening program was effective? Test using αα=.05 and explain the practical conclusion from your test.
d. Find the p-value for the test and interpret it.

Answer:

A) Alternative hypothesis; Ha: p_o > ⅔

B) Null hypothesis;

H0: p_o = ⅔

C) There is sufficient evidence to support the claim that the community screening programme was effective

D) p-value = 0

Its less than the significance value and so we will reject the null hypothesis and conclude that the community screening programme was effective

Step-by-step explanation:

A) We are told that ⅓ of those diagnosed eventually die of the disease. Thus, ⅔ is the fraction that will survive the disease.

Thus, the null hypothesis is;

H0: p_o = ⅔

We are told that a screening programme was started to increase the survival rate. Thus,the alternative hypothesis is;

Ha: p_o > ⅔

B) null hypothesis is;

H0: p_o = ⅔

C) 164 out of the 200 women selected survived the disease.

Thus sample proportion is;

p^ = 164/200

p^ = 0.82

For the z-score value, we will use the formula;

z = (p^ - p_o)/√((p_o(1 - p_o)/n)

Now, p_o = ⅔ = 0.67

Thus;

z = (0.82 - 0.67)/√((0.67(1 - 0.67)/200)

z = 4.51

From online p-value from z-score calculator, we have p-value ≈ 0

This is less than the significant level of 0.05 and therefore we will reject the null hypothesis and conclude that the community screening program was effective.

D) p-value ≈ 0

Reject the null hypothesis

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