Question

A team of eye surgeons has developed a new technique for risky eye operations. Under the old method, it is known that only 30% of the patients who undergo this operation recover their eyesight. Suppose that surgeons in various hospitals have performed a total of 225 operations using the new method and that 88 have been successful. Can we justify the claim that the new method is better than the old one at 1% level of significance?

STEPS:

1. Population characteristics of interest:
2. Null Hypothesis:        H_o:
3. Alternate Hypothesis:   H_a:
4. Significance Level:     a=
5. Test Statistics and its assumptions:

6. Calculation of test statistics:

7. P-Value and Conclusion:

Answer 1

Population characteristics of interest: proportion

Null Hypothesis:
H0: p = 0.3

Alternate Hypothesis:   
Ha: p > 0.3

Significance Level:   
a= 0.01

Test Statistics and its assumptions:
z-statistic
samples are taken from normally distributed population.

Calculation of test statistics:
pcap = 88/225 = 0.3911

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.3911 - 0.3)/sqrt(0.3*0.7/225)
z = 2.98

P-Value and Conclusion:
p-value = 0.0014

p-value < 0.01, reject H0
There is sufficient evidence to conclude that the new method is better.