Identify the p-value and identify the conclusion for this hypothesis test

Question

Consider a drug testing company that provides a test for marijuana usage. Among

303 tested​ subjects, results from

27 subjects were wrong​ (either a false positive or a false​ negative). Use a

0.10 significance level to test the claim that less than

10 percent of the test results are wrong.

Answer 1

The null hypothesis would be that there is no difference between the false negatives and the 10 percent stated, with the alternative being that there is a difference.

H0 Po = Phat

HA Po =/ Phat

We use the proportion of a simple proportion to calculate a test statistic.

z=Phat-Po/sqrt(Po*(1-Po)/n)

Po-.1

Phat=227/303=.08910891

Calculating we get: z=-0.6319335

We then use a program to calculate the P value. I used R.

My code for this is: 2*pnorm(-abs(zscore))

which gives us a p value of 0.5274303.

Given that value, we fail to reject the null hypothesis, meaning there is no difference between the known proporion and the sampled group.

Answer 2

Ho: p0 (population proportion) = 0.1

Ha: p0 < 0.1

This is a single proportion z-test that is left tailed. We will need to compute the z-statistic.

We first need to determine if the test is appropriate. We will assume a random sample. It is safe to assume the 303 is less than 10% of the total number of marajuana users, so the assumption of independence is correct. To assume normality, the large counts condition must be satisfied: > 10 "successes" and > 10 "failures". Here a wrong result is considered a "success". There were 27 "successes" (wrong result) and 303-27 "failures" so the large counts condition is satisfied and we can assume normality and can go forward with the test.

To compute the z-statistic, first we will need to compute the sample proportion (phat) which will equal the number of wrong results (27) / sample size (303).

The z-statistic = (phat - p0)/sqrt(p0*(1-p0)/n), where p0 = 0.1 (the value of p0 under the null hypothesis and n=303).

Since this is a left tailed test, the p-value will be the probability that z < z-statistic.

If that probability is less than 0.1, then we can reject the null hypothesis at the 0.1 level of significance.