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) - 0.07.
B) - 2.33.
C) - 4.12.
D) - 1.86.

Question

Quizplus · Accepted Answer

We can use the one-sample z-test for proportion to test the claim. The null hypothesis is that the proportion of doctors who recommend aspirin is 0.90, and the alternative hypothesis is that the proportion is less than 0.90. The test statistic is calculated as:

z = (p̂ - p) / sqrt(p*(1-p)/n)

where p̂ is the sample proportion, p is the hypothesized proportion under the null hypothesis, and n is the sample size.

In this case, p = 0.90, n = 100, and p̂ = 83/100 = 0.83. Plugging these values into the formula, we get:

z = (0.83 - 0.90) / sqrt(0.90*(1-0.90)/100) = -2.33

Since we are testing the alternative hypothesis that the proportion is less than 0.90, we are interested in the left tail of the standard normal distribution. The p-value can be calculated as the probability of getting a z-value less than or equal to -2.33:

p-value = P(z <= -2.33) = 0.0099

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is evidence to support the alternative hypothesis that the proportion of doctors who recommend aspirin is less than 0.90.

A Survey Claims That 9 Out of 10 Doctors Recommend