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 certain disease. 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 256 operations using the new method, and that 37.5% of the operations have been successful (i.e., the patients fully recovered their sight). Can we justify the claim at the 1% level of significance that the new method has a success rate of more than 30%?

Options:

A. Yes, because the success rate of the new method is 37.5%.

B. No, because the success rate of the old method is 30%.

C. Yes, because the operations using the new method have a higher success rate than the old method.

D. No, because the significance level is too low to draw any conclusion.

Answer 1

Yes we can justify the claim at the 1% level of significance that the new method has a success rate of more than 30%. So, The correct option is C. Yes, because the operations using the new method have a higher success rate than the old method.

Step-by-step explanation:

The claim that the new method has a success rate of more than 30% can be justified at the 1% level of significance. To determine this, we can perform a hypothesis test. Let p represent the true success rate of the new method.

Hypotheses:

- Null hypothesis H₀: p
`\leq` 0.30 (The new method is not more successful than the old method)

- Alternative hypothesis
`H_a` : p > 0.30 (The new method is more successful than the old method)

Test Statistic:

We can use the z-test for proportions. The test statistic z is calculated using the formula:

z =
`\frac{(\hat{p} - p_0)}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}} \]`

where
`\hat{p}` is the sample proportion, p₀ is the assumed proportion under the null hypothesis, and n is the sample size.

Calculation:

Given that
`\hat{p}` = 0.375 , p₀ = 0.30 , and n = 256 , the calculated z-value is obtained.

Decision Rule:

If the calculated z is greater than the critical value for a one-tailed test at a 1% level of significance, we reject the null hypothesis.

Conclusion:

Comparing the calculated z with the critical value, if z >
`z_(\alpha) \)`, we reject H₀ . Therefore, we can justify the claim that the new method has a success rate of more than 30% at the 1% level of significance.