Verified
Given:
n: Total number of patients = 412
x1: Number of females in the sample seeking cancer treatment = 197
x2: Number of males in the sample seeking cancer treatment = 412-197 = 215
Define:
p1: sample proportion of females seeking cancer treatment.
p1= x1/n = 197/412 = 0.4782
p1: sample proportion of males seeking cancer treatment.
p2 = x2/n = 215/412 = 0.5218
We know that the sample proportion is an unbiased estimator of population proportion, hence the proportion of females seeking treatment for skin cancer is
p ^= p1=0.4782
In case of proportions, the population variance is estimated by:
The standard error is nothing but the square root of variance
Hence,
The 95% confidence interval for the population proportion of females seeing treatment for skin cancer is given by:
Where, ZC (Critical value) = 1.96
Hence,
This is the required confidence interval.
Now we are supposed to test whether males are more likely to seek treatment for skin cancer.
Null hypothesis:
H0: There is no significant difference in number of cancer patients due to according to gender
H0: P1 = P2
Alternative hypothesis:
Ha: Males are more likely to seek treatment for skin cancer.
Ha: P1 < P2
α (level of significance) = 0.05 One tailed test
Zα (Critical value) = -1.64
Assumptions:
The two samples come from independent population.
Population is normally distributed.
Test Statistic:
Where,
P = 1/2
Q= 1/2
Hence Z = - 1.2541
P value = P (Z < Z observed)
= P (Z < -1.2541 )
= 0.1049
Decision Rule:
Reject H0 if P value is less than the level of significance.
Decision:
Since observed value (-1.2541) > critical value (-1.64) and P value (observed level of significance) = 0.1049 is greater than α (level of significance) = .05, we fail to reject H0.
Conclusion:
There is no significant difference in number of cancer patients due to according to gender.