In Australia, the overall five-year survival rate for breast cancer in females i
ID: 3072009 • Letter: I
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)
Explanation / Answer
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: Ha: The five-year survival rate in the rural region is not 90%.
H0: p = 0.90 vs. Ha: p 0.90 (Two tailed test)
We are given
Level of significance = = 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 > = 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 = = 0.01
So, confidence level = c = 1 – = 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%.