### Chapter 7. Hypothesis Testing: Hypothesis Test for a Population Proportion

### Large-sample Proportion Test: Test Statistic and p-value

Suppose we draw a simple random sample of size #n# from a population of which some proportion #\pi# exhibits a particular characteristic.

Let #X# denote the number of successes out of #n# observations, and let #\hat{p}=\frac{X}{n}# denote the sample proportion.

When the sample size is *large *(#n\geq 30#), the *Central Limit Theorem *can be invoked, and a #Z#*-test* can be used to test certain hypotheses about the population proportion #\pi#.

#\phantom{0}#

Z-test for a Population Proportion: Test Statistic

The test statistic of a #Z#-test for a population proportion #\pi# is calculated as follows:

\[Z=\cfrac{\hat{p}-\pi_0}{\sigma_{\hat{p}}}=\cfrac{\hat{p}-\pi_0}{\sqrt{\pi_0 \cdot (1 - \pi_0)/n}}\]

Under the assumption that the null hypothesis #H_0# is true, the sampling distribution of #Z# is the *Standard Normal Distribution*. That is, #Z \sim N(0,1)#.

A lowercase #z# will be used to denote the measured value of #Z# after the data has been collected.

Calculating the p-value of a Z-test for a Population Proportion with Statistical Software

The calculation of the #p#-value of a #Z#-test for #\pi# is dependent on the *direction *of the test and can be performed using either Excel or R.

To calculate the #p#-value of a #Z#-test for #\pi# in **Excel**, make use of one of the following commands:

\[\begin{array}{lllll}

\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000}p\text{-value}&\phantom{0000000000}\text{Excel Command}\\

\hline

\text{Two-tailed}&H_0:\pi = \pi_0&H_a:\pi \neq \pi_0&2\cdot \mathbb{P}(Z\geq |z|)&=2 \text{ * }(1 \text{ - }\text{NORM.DIST}(\text{ABS}(z),0,1,1))\\

\text{Left-tailed}&H_0:\pi \geq \pi_0&H_a:\pi \lt \pi_0&\mathbb{P}(Z\leq z)&=\text{NORM.DIST}(z,0,1,1)\\

\text{Right-tailed}&H_0:\pi \leq \pi_0&H_a:\pi \gt \pi_0&\mathbb{P}(Z\geq z)&=1 \text{ - }\text{NORM.DIST}(z,0,1,1)\\

\end{array}\]

To calculate the #p#-value of a #Z#-test for #\pi# in **R**, make use of one of the following commands:

#\begin{array}{lllll}

\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000}p\text{-value}&\phantom{000000}\text{R Command}\\

\hline

\text{Two-tailed}&H_0:\pi = \pi_0&H_a:\pi \neq \pi_0&2\cdot \mathbb{P}(Z\geq |z|)&2 \text{ * }\text{pnorm}(\text{abs}(z),0,1, \text{FALSE})\\

\text{Left-tailed}&H_0:\pi \geq \pi_0&H_a:\pi \lt \pi_0&\mathbb{P}(Z\leq z)&\text{pnorm}(z,0,1, \text{TRUE})\\

\text{Right-tailed}&H_0:\pi \leq \pi_0&H_a:\pi \gt \pi_0&\mathbb{P}(Z\geq z)&\text{pnorm}(z,0,1, \text{FALSE})\\

\end{array}#

If #p \leq \alpha#, reject #H_0# and conclude #H_a#. Otherwise, do not reject #H_0#.

Note: #|z|# denotes the *absolute value *of #z#, i.e. the distance of #z# from #0#, regardless of whether #z < 0# or #z > 0#.

To test this claim, a journalist working at a newspaper surveys #100# citizens, using random sampling. Let #X# denote the number of citizens who are in favor of the new law.

The journalist plans on using a hypothesis test to determine if the proportion of citizens who are in favor of the new law significantly

*differs*from #0.85#, at the #\alpha = 0.07# level of significance.

The survey results indicate that #81# of the citizens in the sample are in favor of the new proposal.

Calculate the #p#-value of the test and make a decision regarding #H_0: \pi = 0.85#. Round your answer to #4# decimal places.

On the basis of this #p#-value, #H_0# should not be rejected, because #\,p# #\gt# #\alpha#.

There are a number of different ways we can calculate the #p#-value of the test. Click on one of the panels to toggle a specific solution.

A sample size of #100# is considered large enough for the *Central Limit Theorem *to apply. This means that the test statistic

\[Z=\cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0 \cdot (1-\pi_0) / n}}\]

approximately has the #N(0,1)# distribution under the assumption that #H_0# is true.

Compute the sample proportion #\hat{p}#:

\[\hat{p} = \cfrac{X}{n} = \cfrac{81}{100} = 0.81000\]

Calculate the value of *test statistic* #z#:

\[z = \cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0\cdot(1-\pi_0)/n}} = \cfrac{0.81000 - 0.85}{\sqrt{0.85\cdot(1 - 0.85) / 100}} = -1.12022\]

For a *two-tailed *#Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in Excel, make use of the following function:

NORM.DIST(x, mean, standard_dev, cumulative)

x: The value at which you wish to evaluate the distribution function.mean: The mean of the distribution.standard_dev: The standard deviation of the distribution.cumulative: A logical value that determines the form of the function.

- TRUE - uses the cumulative distribution function, #\mathbb{P}(X \leq x)#
- FALSE - uses the probability density function

Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:

\[

=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(z),0,1,1))\\

\downarrow\\

=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(\text{-}1.12022),0,1,1))

\]

This gives:

\[p = 0.2626\]

Since #\,p# #\gt# #\alpha#, #H_0: \pi = 0.85# should not be rejected.

A sample size of #100# is considered large enough for the *Central Limit Theorem *to apply. This means that the test statistic

\[Z=\cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0 \cdot (1-\pi_0) / n}}\]

approximately has the #N(0,1)# distribution under the assumption that #H_0# is true.

Compute the sample proportion #\hat{p}#:

\[\hat{p} = \cfrac{X}{n} = \cfrac{81}{100} = 0.81000\]

Calculate the value of *test statistic* #z#:

\[z = \cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0\cdot(1-\pi_0)/n}} = \cfrac{0.81000 - 0.85}{\sqrt{0.85\cdot(1 - 0.85) / 100}} = -1.12022\]

For a *two-tailed *#Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in R, make use of the following function:

pnorm(q, mean, sd, lower.tail)

q: The value at which you wish to evaluate the distribution function.mean: The mean of the distribution.sd: The standard deviation of the distribution.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:

\[

2 \text{ * } \text{pnorm}(q = \text{abs}(z), mean = 0, sd = 1,lower.tail = \text{FALSE})\\

\downarrow\\

2\text{ * } \text{pnorm}(q = \text{abs}(\text{-}1.12022), mean = 0, sd = 1,lower.tail = \text{FALSE})

\]

This gives:

\[p = 0.2626\]

Since #\,p# #\gt# #\alpha#, #H_0: \pi = 0.85# should not be rejected.

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