Final answer:
Based on the sample of 250 lab reports with 85 errors, hypothesis testing at a 0.05 significance level indicates sufficient evidence to support the hospital director's claim that over 30% of the reports contain errors, as the p-value is less than 0.05, leading to rejection of the null hypothesis.
Step-by-step explanation:
The student's question revolves around testing a statistical claim using hypothesis testing.
To determine whether the hospital director's belief that more than 30% of lab reports contain errors is substantiated, one would use a one-sample z-test for proportions. Here's how the hypothesis test is set up:
- The null hypothesis (H0) would state that the proportion of errors in lab reports is 30% or less.
- The alternative hypothesis (H1) would claim that the proportion of errors is greater than 30%.
Given a sample size of 250 reports with 85 errors, the sample proportion is 85/250 = 0.34.
At a significance level of alpha: 0.05,
we perform the test to see if this provides enough statistical evidence to reject the null hypothesis.
If the calculated p-value is less than 0.05, the decision will be to reject the null hypothesis, which means there is sufficient evidence at the 5 percent significance level to suggest that the proportion of errors in the lab reports is indeed greater than 30%.
In this case, the student's hypothetical results indicate that the p-value is less than 0.05, hence, the conclusion is that there's sufficient evidence to support the hospital director's claim.