Question

2. (4pts) It has been reported that one in four medical doctors received their

degrees from foreign schools. A hospital researcher believes that the
percentage is less than 25% A random survey of 100 medical doctors found
that 12 received their degrees from a foreign school. At a = 0.05, is there
enough evidence to support the researcher's claim.

Answer 1

Okay, let's go through this step-by-step:

The claim is that the percentage of medical doctors who received their degrees from foreign schools is less than 25%.

The null hypothesis is that the percentage is equal to 25% (the reported percentage).

The alternative hypothesis is that the percentage is less than 25%.

n = 100 (the number of doctors surveyed)

x = 12 (the number of doctors in the sample who received foreign degrees)

The test statistic is:

z = (x - np) / √(np(1-p))

Where:

n = sample size

p = population proportion under null hypothesis

x = number of successes in sample

Plugging in the values:

z = (12 - 100*(0.25)) / √(100*(0.25)*(1-0.25))

z = -2.12

Using a significance level (α) of 0.05, the critical value is -1.64 (from a z-table).

Since the test statistic of -2.12 is less than the critical value of -1.64, we reject the null hypothesis.

Therefore, there is enough evidence at the 5% significance level to support the researcher's claim that the percentage of doctors with foreign degrees is less than 25%.