A survey claims that 9 out of 10 doctors (i.e., 90%) recommend brand Z for their patients who have children. Totest this claim against the alternative that the actual proportion of doctors who recommend brand Z is less than90%, a random sample of 100 doctors results in 83 who indicate that they recommend brand Z. The test statisticin this problem is approximately (round to the nearest hundredth):
A) -2.33
B) 2.33
C) -1.83
D) -1.99

Question

Quizplus · Accepted Answer

The test statistic is calculated using the formula for a one-sample proportion z-test: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$, where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size. Here, $\hat{p} = \frac{83}{100} = 0.83$, $p_0 = 0.90$, and $n = 100$. Plugging these values into the formula gives $z \approx -2.33$, indicating that the sample proportion of doctors who recommend brand Z is significantly lower than the claimed proportion, assuming a normal distribution.

A Survey Claims That 9 Out of 10 Doctors (I