A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors' results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to:
A) -1.67
B) -2.33
C) -1.86
D) -0.14
E) -2.58

Question

Quizplus · Accepted Answer

The null hypothesis is that 90% of doctors recommend aspirin, so the probability of a doctor recommending aspirin is 0.9. The alternative hypothesis is that the proportion of doctors who recommend aspirin is less than 0.90.

The sample size is n=100, the number of doctors who recommend aspirin is x=83. So, the sample proportion is $\hat p = \frac{x}{n} = \frac{83}{100} = 0.83$.

Under the null hypothesis, the test statistic is approximately standard normal, so we can use the standard normal distribution to find the p-value. The test statistic is:

$z = \frac{\hat p - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.83-0.90}{\sqrt{\frac{0.9	imes0.1}{100}}} \approx -2.33$

Using a standard normal table or calculator, the p-value is approximately 0.0107. Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the actual proportion of doctors who recommend aspirin is less than 0.90.

The best choice is B because it is the closest approximation of the test statistic.

A Survey Claims That 9 Out of 10 Doctors Recommend