Question

A team of eye surgeons has developed a new technique for a risky eye operation to restore the sight of people blinded by a rare disease. Under the current, old method of operation, it is known that 30% of the patients that undergo the surgery regain their eyesight.

The surgeons complete 225 operations using the new method, with 84 of the surgeries resulting in the restoration of eyesight in the patients. Can the surgeons claim that the new method is better than the old method? Test the results at the 5% level of significance and the 1% level of significance.

For the null hypothesis assume that the new method is not different from the old method. For the null hypothesis, use the proportion from the old method.

(a) Calculate the sample proportion: =

(b) State the null and the alternative hypothesis.

      H0 : p =

           

Ha : p [ ≠ or > or < ]

(c) Does the sample size satisfy the Central Limit Theorem?

(d) Should the test be a left-tail, right-tail or two- tail test?

(e) Calculate z using the formula above.

(f) Find the p-value using the standard normal distribution.

(g) Decide whether to reject the null hypothesis and explain what it means in context.

           

Decision at significance level = 0.01

        

      Base on the significance level 0.01 is the new method the same or better than the old method?

        

Decision at significance level = 0.05

Base on the significance level 0.05 is the new method the same or better than the old method?

Answer 1

Ans:

a)

sample proportion=84/225=0.3733

b)

H0 : p =0.30           

Ha : p >0.30

c)Yes,as np and nq>5

d)Right tailed

e)Test statistic:

z=(0.3733-0.3)/SQRT(0.3*(1-0.3)/225)

z=2.40

f)p-value=P(z>2.40)=0.0082

g)

As,p-value<0.01,we reject the null hypothesis and we can conclude that new method is better than the old method.

As,p-value<0.05,we reject the null hypothesis and we can conclude that new method is better than the old method.