Question

Ti-84

A research group claims that less than 25% of students at one medical school plan to go into general practice. It is found that among a random sample of 120 of the school's students, 20% of them plan to go into general practice. At the 0.10 significance level, do the data provide sufficient evidence to conclude that the percentage of all students at this school who plan to go into general practice is less than 25%? Use the confidence interval approach.

Answer 1

The above test is a one-sided test with the following Null and Alternative Hypotheses :

Now, using sample data, we calculate the standard deviation (σ) and compute the z-score test statistic (z) as :

Note : Here, is the hypothesized value of population proportion in the null hypothesis, is the sample proportion, and is the sample size.

Since, we have a one tailed test, the P-value is the probability that the z-score is less than -1.26, that is, we are going to obtain the value of the probability of .

Thus, the p-value is obtained as :

Now the given level of significance is 0.1. Thus, we observe that P-value (0.1) is equal to the significance level (0.1), hence we do not accept the null hypothesis.

Thus, we can conclude that the percentage of all students at this school who plan to go into general practice is less than 25%, under the light of the given sample.

Thus, the confidence interval is obtained as :

where is the value of for which the area under the curve of a normal distribution with mean 0 and standard deviation 1 is equal to 0.1.

Thus, the confidence interval is obtained as :

Thus, the confidence interval for the proportion of all students at this school who plan to go into general practice is obtained as at 0.1 level of significance.