Statistics - One Proportion Z Test
The test statistic is a z-score (z) defined by the following equation. ${z = \frac{(p - P)}{\sigma}}$ where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and ${\sigma}$ is the standard deviation of the sampling distribution.
Test Statistics is defined and given by the following function:
Formula
${ z = \frac {\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}} }$
Where −
Example
Problem Statement:
A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim, a random sample of 100 doctors is obtained. Of these 100 doctors, 82 indicate that they recommend aspirin. Is this claim accurate? Use alpha = 0.05.
Solution:
Define Null and Alternative Hypotheses
${ H_0;p = .90 \\[7pt]
H_0;p \ne .90 }$
Here Alpha = 0.05. Using an alpha of 0.05 with a two-tailed test, we would expect our distribution to look something like this:
Here we have 0.025 in each tail. Looking up 1 - 0.025 in our z-table, we find a critical value of 1.96. Thus, our decision rule for this two-tailed test is: If Z is less than -1.96, or greater than 1.96, reject the null hypothesis.Calculate Test Statistic:
${ z = \frac {\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}} \\[7pt]
\hat p = .82 \\[7pt]
p_o = .90 \\[7pt]
n = 100 \\[7pt]
z_o = \frac {.82 - .90}{\sqrt{\frac{ .90 (1- .90)}{100}}} \\[7pt]
\ = \frac{-.08}{0.03} \\[7pt]
\ = -2.667 }$
As z = -2.667 Thus as result we should reject the null hypothesis and as conclusion, The claim that 9 out of 10 doctors recommend aspirin for their patients is not accurate, z = -2.667, p < 0.05.
