# 8.5: Large Sample Tests for a Population Proportion

Skills to Develop

• To learn how to apply the five-step critical value test procedure for test of hypotheses concerning a population proportion.
• To learn how to apply the five-step $$p$$-value test procedure for test of hypotheses concerning a population proportion.

Both the critical value approach and the p-value approach can be applied to test hypotheses about a population proportion p. The null hypothesis will have the form $$H_0 : p = p_0$$ for some specific number $$p_0$$ between $$0$$ and $$1$$. The alternative hypothesis will be one of the three inequalities

• $$p <p_0$$,
• $$p>p_0$$, or
• $$p≠p_0$$

for the same number $$p_0$$ that appears in the null hypothesis.

The information in Section 6.3 gives the following formula for the test statistic and its distribution. In the formula $$p_0$$ is the numerical value of $$p$$ that appears in the two hypotheses, $$q_0=1−p_0, \hat{p}$$ is the sample proportion, and $$n$$ is the sample size. Remember that the condition that the sample be large is not that $$n$$ be at least 30 but that the interval

$\left[ \hat{p} −3 \sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} , \hat{p} + 3 \sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} \right]$

lie wholly within the interval $$[0,1]$$.

Standardized Test Statistic for Large Sample Hypothesis Tests Concerning a Single Population Proportion

$Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0q_o}{n}}} \label{eq2}$

The test statistic has the standard normal distribution.

The distribution of the standardized test statistic and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure $$\PageIndex{1}$$.

Figure $$\PageIndex{1}$$: Distribution of the Standardized Test Statistic and the Rejection Region

Example $$\PageIndex{1}$$

A soft drink maker claims that a majority of adults prefer its leading beverage over that of its main competitor’s. To test this claim $$500$$ randomly selected people were given the two beverages in random order to taste. Among them, $$270$$ preferred the soft drink maker’s brand, $$211$$ preferred the competitor’s brand, and $$19$$ could not make up their minds. Determine whether there is sufficient evidence, at the $$5\%$$ level of significance, to support the soft drink maker’s claim against the default that the population is evenly split in its preference.

Solution:

We will use the critical value approach to perform the test. The same test will be performed using the $$p$$-value approach in Example $$\PageIndex{3}$$.

We must check that the sample is sufficiently large to validly perform the test. Since $$\hat{p} =270/500=0.54$$,

$\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} =\sqrt{ \dfrac{(0.54)(0.46)}{500}} ≈0.02$

hence

\begin{align} & \left[ \hat{p} −3\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} ,\hat{p} +3\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} \right] \\ &=[0.54−(3)(0.02),0.54+(3)(0.02)] \\ &=[0.48, 0.60] ⊂[0,1] \end{align}

so the sample is sufficiently large.

• Step 1. The relevant test is

$H_0 : p = 0.50$

$vs.$

$H_a : p > 0.50\, @ \,\alpha =0.05$

where $$p$$ denotes the proportion of all adults who prefer the company’s beverage over that of its competitor’s beverage.

• Step 2. The test statistic (Equation \ref{eq2}) is
$Z=\dfrac{\hat{p} −p_0}{\sqrt{ \dfrac{p_0q_0}{n}}}$

and has the standard normal distribution.

• Step 3. The value of the test statistic is
\begin{align} Z &=\dfrac{\hat{p} −p_0}{\sqrt{ \dfrac{p_0q_0}{n}}} \\[6pt] &= \dfrac{0.54−0.50}{\sqrt{\dfrac{(0.50)(0.50)}{500}}} \\[6pt] &=1.789 \end{align}
• Step 4. Since the symbol in $$H_a$$ is “$$>$$” this is a right-tailed test, so there is a single critical value, $$z_{α}=z_{0.05}$$. Reading from the last line in Figure 7.1.6 its value is $$1.645$$. The rejection region is $$[1.645,∞)$$.
• Step 5. As shown in Figure $$\PageIndex{2}$$ the test statistic falls in the rejection region. The decision is to reject $$H_0$$. In the context of the problem our conclusion is:

The data provide sufficient evidence, at the $$5\%$$ level of significance, to conclude that a majority of adults prefer the company’s beverage to that of their competitor’s.

Figure $$\PageIndex{2}$$: Rejection Region and Test Statistic for Example $$\PageIndex{1}$$

Example $$\PageIndex{2}$$

Globally the long-term proportion of newborns who are male is $$51.46\%$$. A researcher believes that the proportion of boys at birth changes under severe economic conditions. To test this belief randomly selected birth records of $$5,000$$ babies born during a period of economic recession were examined. It was found in the sample that $$52.55\%$$ of the newborns were boys. Determine whether there is sufficient evidence, at the $$10\%$$ level of significance, to support the researcher’s belief.

Solution:

We will use the critical value approach to perform the test. The same test will be performed using the $$p$$-value approach in Example $$\PageIndex{1}$$.

The sample is sufficiently large to validly perform the test since

$\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} =\sqrt{ \dfrac{(0.5255)(0.4745)}{5000}} ≈0.01$

hence

\begin{align} & \left[ \hat{p} −3\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} ,\hat{p} +3\sqrt{ \dfrac{\hat{p} (1−\hat{p} )}{n}} \right] \\ &=[0.5255−0.03,0.5255+0.03] \\ &=[0.4955,0.5555] ⊂[0,1] \end{align}
• Step 1. Let $$p$$ be the true proportion of boys among all newborns during the recession period. The burden of proof is to show that severe economic conditions change it from the historic long-term value of $$0.5146$$ rather than to show that it stays the same, so the hypothesis test is

$H_0 : p = 0.5146$

$vs.$

$H_a : p \neq 0.5146\, @ \,\alpha =0.10$

• Step 2. The test statistic (Equation \ref{eq2}) is

$Z=\dfrac{\hat{p} −p_0}{\sqrt{ \dfrac{p_0q_0}{n}}}$

and has the standard normal distribution.

• Step 3. The value of the test statistic is
\begin{align} Z &=\dfrac{\hat{p} −p_0}{\sqrt{ \dfrac{p_0q_0}{n}}} \\[6pt] &= \dfrac{0.5255−0.5146}{\sqrt{\dfrac{(0.5146)(0.4854)}{5000}}} \\[6pt] &=1.542 \end{align}
• Step 4. Since the symbol in $$H_a$$ is “$$\neq$$” this is a two-tailed test, so there are a pair of critical values, $$\pm z_{\alpha /2}=\pm z_{0.05}=\pm 1.645$$. The rejection region is $$(-\infty ,-1.645]\cup [1.645,\infty )$$.
• Step 5. As shown in Figure $$\PageIndex{3}$$ the test statistic does not fall in the rejection region. The decision is not to reject $$H_0$$. In the context of the problem our conclusion is:

The data do not provide sufficient evidence, at the $$10\%$$ level of significance, to conclude that the proportion of newborns who are male differs from the historic proportion in times of economic recession.

Figure $$\PageIndex{3}$$: Rejection Region and Test Statistic for Example $$\PageIndex{2}$$.

Example $$\PageIndex{3}$$

Perform the test of Example $$\PageIndex{1}$$ using the $$p$$-value approach.

Solution:

We already know that the sample size is sufficiently large to validly perform the test.

• Steps 1–3 of the five-step procedure described in Section 8.3 have already been done in Example $$\PageIndex{1}$$ so we will not repeat them here, but only say that we know that the test is right-tailed and that value of the test statistic is $$Z = 1.789$$.
• Step 4. Since the test is right-tailed the p-value is the area under the standard normal curve cut off by the observed test statistic, $$Z = 1.789$$, as illustrated in Figure $$\PageIndex{4}$$. By Figure 7.1.5 that area and therefore the p-value is $$1−0.9633=0.0367$$.
• Step 5. Since the $$p$$-value is less than $$α=0.05$$ the decision is to reject $$H_0$$.

Figure $$\PageIndex{4}$$: P-Value for Example $$\PageIndex{3}$$

Example $$\PageIndex{4}$$

Perform the test of Example $$\PageIndex{2}$$ using the $$p$$-value approach.

Solution:

We already know that the sample size is sufficiently large to validly perform the test.

• Steps 1–3 of the five-step procedure described in Section 8.3 have already been done in Example $$\PageIndex{2}$$. They tell us that the test is two-tailed and that value of the test statistic is $$Z = 1.542$$.
• Step 4. Since the test is two-tailed the $$p$$-value is the double of the area under the standard normal curve cut off by the observed test statistic, $$Z = 1.542$$. By Figure 7.1.5 that area is $$1-0.9382=0.0618$$, as illustrated in Figure $$\PageIndex{5}$$, hence the $$p$$-value is $$2\times 0.0618=0.1236$$.
• Step 5. Since the $$p$$-value is greater than $$\alpha =0.10$$ the decision is not to reject $$H_0$$.

Figure $$\PageIndex{5}$$: P-Value for Example $$\PageIndex{4}$$

Key Takeaway

• There is one formula for the test statistic in testing hypotheses about a population proportion. The test statistic follows the standard normal distribution.
• Either five-step procedure, critical value or $$p$$-value approach, can be used.

• Anonymous