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%.