Question

In Australia, the overall five-year survival rate for breast cancer in females is 90%. A random sample of 420 breast cancer patients in a rural region was randomly selected and followed up after 5 years. The results show that 365 of them survived after the 5 years of follow up period. Four students used these results as a hypothesis test to indicate whether the proportion of survival in this rural region is not the same as the national rate of 90%, using a 1% level of significance. Choose the correct calculation ( Note: we normally would prefer to obtain the confidence interval for the estimated proportion and the associated p-value, which provide the level of evidence against the null hypothesis.)

Select one:

a. Reject the H₀ (z=-2.11)

b. Reject the H0 (z=2.11)

c. Not reject the H0 (z=2.11)

d. Not reject the H0 (z=-2.11)

Answer 1

Correct Answer: d. Not reject the H0 (z=-2.11)

Solution:

Here, we have to use the z-test for population proportion and confidence interval for population proportion for checking given hypothesis or claim. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The five-year survival rate in the rural region is 90%.

Alternative hypothesis: H_a: The five-year survival rate in the rural region is not 90%.

H₀: p = 0.90 vs. H_a: p ≠ 0.90 (Two tailed test)

We are given

Level of significance = α = 0.01

Sample size = n = 420

Number of favourable observations = X = 365

Sample proportion = P = X/n = 365/420 = 0.869047619

Test statistic is given as below:

Z = (P – p) / sqrt(p*(1 – p)/n)

Z = (0.869047619 - 0.90) / sqrt(0.90*(1 - 0.90)/420)

Z = (0.869047619 - 0.90) /0.0146

Z = -2.11445

Critical value = -2.5758 and 2.5758 (by using z-table)

P-value = 0.0345 (by using z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis that the five year survival rate in rural region is 90%.

There is insufficient evidence to conclude that the proportion of survival in this rural region is not the same as the national rate of 90%.

We are given

Level of significance = α = 0.01

So, confidence level = c = 1 – α = 1 – 0.01 = 0.99

Now, we have to find 99% confidence interval for population proportion.

Confidence interval = P ± Z*sqrt(P*(1 – P)/n)

Critical Z value = 2.5758 (by using z-table)

Confidence interval = 0.869047619 ± 2.5758*sqrt(0.869047619*(1 - 0.869047619)/420)

Confidence interval = 0.869047619 ± 2.5758* 0.0165

Confidence interval = 0.869047619 ± 0.0424

Lower limit = 0.869047619 - 0.0424 = 0.8266

Upper limit = 0.869047619 + 0.0424 = 0.9114

Confidence interval = (0.8266, 0.9114)

The value for claimed proportion of 0.90 is included in the given confidence interval, so we do not reject the null hypothesis.

There is insufficient evidence to conclude that the proportion of survival in this rural region is not the same as the national rate of 90%.