8.1.2.1  Normal Approximation Method Formulas
8.1.2.1  Normal Approximation Method FormulasHere we will be using the five step hypothesis testing procedure to compare the proportion in one random sample to a specified population proportion using the normal approximation method.
In order to use the normal approximation method, the assumption is that both \(n p_0 \geq 10\) and \(n (1p_0) \geq 10\). Recall that \(p_0\) is the population proportion in the null hypothesis.
Research Question  Is the proportion different from \(p_0\)?  Is the proportion greater than \(p_0\)?  Is the proportion less than \(p_0\)? 

Null Hypothesis, \(H_{0}\)  \(p=p_0\)  \(p= p_0\)  \(p= p_0\) 
Alternative Hypothesis, \(H_{a}\)  \(p\neq p_0\)  \(p> p_0\)  \(p< p_0\) 
Type of Hypothesis Test  Twotailed, nondirectional  Righttailed, directional  Lefttailed, directional 
Where \(p_0\) is the hypothesized population proportion that you are comparing your sample to.
When using the normal approximation method we will be using a z test statistic. The z test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. Note that this formula follows the basic structure of a test statistic that you learned last week: \(test\;statistic=\frac{sample\;statisticnull\;parameter}{standard\;error}\)
 Test statistic: One Group Proportion
 \(z=\frac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}\)

\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size
Given that the null hypothesis is true, the p value is the probability that a randomly selected sample of n would have a sample proportion as different, or more different, than the one in our sample, in the direction of the alternative hypothesis. We can find the p value by mapping the test statistic from step 2 onto the z distribution.
Note that pvalues are also symbolized by \(p\). Do not confuse this with the population proportion which shares the same symbol.
We can look up the \(p\)value using Minitab Express by constructing the sampling distribution. Because we are using the normal approximation here, we have a \(z\) test statistic that we can map onto the \(z\) distribution. Recall, the z distribution is a normal distribution with a mean of 0 and standard deviation of 1. If we are conducting a onetailed (i.e., right or lefttailed) test, we look up the area of the sampling distribution that is beyond our test statistic. If we are conducting a twotailed (i.e., nondirectional) test there is one additional step: we need to multiple the area by two to take into account the possibility of being in the right or left tail.
We can decide between the null and alternative hypotheses by examining our pvalue. If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis. Unless stated otherwise, assume that \(\alpha=.05\).
When we reject the null hypothesis our results are said to be statistically significant.
Based on our decision in step 4, we will write a sentence or two concerning our decision in relation to the original research question.
8.1.2.1.1  Video Example: Male Babies
8.1.2.1.1  Video Example: Male Babies8.1.2.1.2  Example Handedness
8.1.2.1.2  Example HandednessResearch Question: Are more than 80% of American's right handed?
In a sample of 100 Americans, 87 were right handed.
\(np_0 = 100(0.80)=80\)
\(n(1p_0) = 100 (10.80) = 20\)
Both \(np_0\) and \(n(1p_0)\) are at least 10 so we can use the normal approximation method.
This is a righttailed test because we want to know if the proportion is greater than 0.80.
\(H_{0}\colon p=0.80\)
\(H_{a}\colon p>0.80\)
 Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}\)
\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size
\(\widehat{p}=\dfrac{87}{100}=0.87\), \(p_{0}=0.80\), \(n=100\)
\(z= \dfrac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}= \dfrac{0.870.80}{\sqrt{\frac{0.80 (10.80)}{100}}}=1.75\)
Our \(z\) test statistic is 1.75.
This is a righttailed test so we need to find the area to the right of the test statistic, \(z=1.75\), on the z distribution.
Using Minitab Express, we find the probability \(P(z\geq1.75)=0.0400592\) which may be rounded to \(p\; value=0.0401\).
\(p\leq .05\), therefore our decision is to reject the null hypothesis
Yes, there is statistical evidence to state that more than 80% of all Americans are right handed.
8.1.2.1.3  Example: Ice Cream
8.1.2.1.3  Example: Ice CreamResearch Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%?
In a sample of 50 customers 60% preferred chocolate over vanilla.
\(np_0 = 50(0.80) = 40\)
\(n(1p_0)=50(10.80) = 10\)
Both \(np_0\) and \(n(1p_0)\) are at least 10. We can use the normal approximation method.
This is a lefttailed test because we want to know if the proportion is less than 0.80.
\(H_{0}\colon p=0.80\)
\(H_{a}\colon p<0.80\)
 Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}\)
\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size
\(\widehat{p}=0.60\), \(p_{0}=0.80\), \(n=50\)
\(z= \dfrac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}= \dfrac{0.600.80}{\sqrt{\frac{0.80 (10.80)}{50}}}=3.536\)
Our \(z\) test statistic is 3.536.
This is a lefttailed test so we need to find the area to the right of our test statistic, \(z=3.536\).
From the Minitab Express output above, the pvalue is 0.0002031
\(p \leq.05\), therefore our decision is to reject the null hypothesis.
Yes, there is evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.
8.1.2.1.4  Example: Overweight Citizens
8.1.2.1.4  Example: Overweight CitizensAccording to the Center for Disease Control (CDC), the percent of adults 20 years of age and over in the United States who are overweight is 69.0% (see http://www.cdc.gov/nchs/fastats/obesityoverweight.htm). One city’s council wants to know if the proportion of overweight citizens in their city is different from this known national proportion. They take a random sample of 150 adults 20 years of age or older in their city and find that 98 are classified as overweight. Let’s use the five step hypothesis testing procedure to determine if there is evidence that the proportion in this city is different from the known national proportion.
\(np_0 =150 (0.690)=103.5 \)
\(n (1p_0) =150 (10.690)=46.5\)
Both \(n p_0\) and \(n (1p_0)\) are at least 10, this assumption has been met.
Research question: Is this city’s proportion of overweight individuals different from 0.690?
This is a nondirectional test because our question states that we are looking for a differences as opposed to a specific direction. This will be a twotailed test.
\(H_{0}\colon p=0.690\)
\(H_{a}\colon p\neq 0.690\)
 Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p} p_0 }{\sqrt{\frac{p_0 (1 p_0)}{n}}}\)
\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size
\(\widehat{p}=\dfrac{98}{150}=.653\)
\( z =\dfrac{0.653 0.690 }{\sqrt{\frac{0.690 (1 0.690)}{150}}} = 0.980 \)
Our test statistic is \(z=0.980\)
This is a nondirectional (i.e., twotailed) test, so we need to find the area under the z distribution that is more extreme than \(z=0.980\).
In Minitab Express, we find the proportion of a normal curve beyond \(\pm0.980\):
\(pvalue=0.163543+0.163543=0.327086\)
\(p>\alpha\), therefore we fail to reject the null hypothesis
There is not sufficient evidence to state that the proportion of citizens of this city who are overweight is different from the national proportion of 0.690.