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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 −
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${z}$ = Test statistics
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${n}$ = Sample size
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${p_o}$ = Null hypothesized value
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${hat p}$ = Observed proportion
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.
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