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

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 $\pi$ .
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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 $p \lt \alpha$ , 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 $z > 0$ .

A politician claims that $65\%$ of citizens are in favor of her new law proposal.

To test this claim, a journalist working at a newspaper surveys $105$ 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 $0.65$ , at the $\alpha = 0.04$ level of significance.

The survey results indicate that $69$ 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 $H_0: \pi = 0.65$ . Round your answer to $4$ decimal places.

$p=0.8780$

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.

Excel Calculation

A sample size of $105$ 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 $\hat{p}$ :
$\hat{p} = \cfrac{X}{n} = \cfrac{69}{105} = 0.65714$
Calculate the value of test statistic $z$ :
$z = \cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0\cdot(1-\pi_0)/n}} = \cfrac{0.65714 - 0.65}{\sqrt{0.65\cdot(1 - 0.65) / 105}} = 0.15345$
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:
$=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(z),0,1,1))\\ \downarrow\\ =2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(0.15345),0,1,1))$
This gives:
$p = 0.8780$
Since $\,p$ $\gt$ $\alpha$ , $H_0: \pi = 0.65$ should not be rejected.

R Calculation

A sample size of $105$ is considered large enough for the Central Limit Theorem to apply. This means that the test statistic
$Z=\cfrac{\hat{p} - \pi_0}{\sqrt{\pi_0 \cdot (1-\pi_0) / n}}$

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:
$2 \text{ * } \text{pnorm}(q = \text{abs}(z), mean = 0, sd = 1,lower.tail = \text{FALSE})\\ \downarrow\\ 2\text{ * } \text{pnorm}(q = \text{abs}(0.15345), mean = 0, sd = 1,lower.tail = \text{FALSE})$
This gives:
$p = 0.8780$
Since $\,p$ $\gt$ $\alpha$ , $H_0: \pi = 0.65$ should not be rejected.

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