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QUAN 2010 Notes: Introduction to Business Statistics

Omar Ortega

Section 10.5 Hypothesis Testing for the Population Proportion

Now we want to see how we can use information about a sample proportion to see what we can conclude about the population proportion.

Investigation 10.5.1.

A Harris poll surveyed 2016 adults and found that 463 of them had one or more tattoos. Does the sample provide evidence that the percentage of adults who have a tattoo is less than 25%?
Figure 10.5.1.
Tattoo Simulation (Made in GeoGebra by Steve Phelps)
The last type of hypothesis test that we will study in this chapter involves the population proportion, \(p\text{.}\) This test is useful for testing claims made about the proportion of something. Recall that proportion data follow the binomial distribution, which can be approximated by the normal distribution under the conditions:
\begin{equation*} np\geq 5 \text{ and } n(1-p)\geq 5,\text{ where} \end{equation*}
\(p=\) the probability of success in the population
\(n=\) the sample size
When these conditions are met, the test statistic for this hypothesis test is calculated with this formula:
\begin{equation*} z_p = \frac{\overline{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \end{equation*}

Exercise 10.5.2.

(Donnelly 9.31)
In April 2010, \(45\%\) of the unemployed had been out of work longer than six months. Policy makers felt that this percentage declined during 2010 as the job market improved. To test this theory, a random sample of 200 unemployed people was selected, and it was found that 80 had been out of work for more than six months. Assuming \(\alpha = 0.10\text{,}\) what conclusions can be drawn about the proportion of the unemployed who have been out of work for more than six months? Use the p-value method of hypothesis testing.
Answer.
\(n=200,\; \alpha=0.10,\; p=.45,\; \overline{p}=\frac{80}{200}\)
\(np=200(.45)=90\geq 5\)
\(n(1-p)=200(.55)\geq 5\)
\(H_0:\; p\geq .45\)
\(H_1:\; p\lt .45\)
\begin{equation*} z_p=\frac{.40-.45}{\sqrt{\frac{.45(.55)}{200}}}\approx -1.42 \end{equation*}
\begin{equation*} \text{p-value}\approx NORM.S.DIST(-1.42,1)\approx 0.0778 \end{equation*}
\begin{equation*} \text{p-value}\lt\alpha\rightarrow \text{ Reject } H_0 \end{equation*}
There is enough evidence to conclude that the proportion of unemployed out of work more than 6 months declined as the job market improved.

Exercise 10.5.3.

(Donnelly 9.26)
An increased number of colleges have been using online resources to research applicants. According to a study from last year, \(33\%\) of admissions officers indicated that they visited an applying student’s social networking page. A random sample of 500 admissions officers was recently selected and it was found that 170 of them visit the social networking sites of students applying to their college. Assuming \(\alpha = 0.05\text{,}\) does this sample provide support for the hypothesis that the proportion of admissions officers who visit an applying students’ social networking page has increased in the past year? Use the traditional method of hypothesis testing.
Answer.
\(n=500,\; p=0.33,\; \overline{p}=\frac{170}{500}=0.34\)
\(H_0: \; p\leq 0.33\)
\(\alpha=0.05\)
\(np=500(.33)\geq 5\)
\(n(1-p)=500(.67)\geq 5\)
\(H_1: \; p\gt 0.33\)
critical value: \(z_{.05}=NORM.S.INV(.95)\approx 1.645\)
test statistic: \(z_p=\frac{.34-.33}{\sqrt{\frac{.33(.67)}{500}}}\approx .48\)
Since \(z_p\lt z_{.05}\) and this is a right-tailed test, the test statistic does not fall in the rejection region, and we fail to reject \(H_0\text{.}\)
The sample does not provide enough evidence to support the claim that the proportion of admissions officials who visit a student’s social networking page has increased in the past year.
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