## 8.8 Hypothesis Tests for a Population Proportion

Learning objectives.

• Conduct and interpret hypothesis tests for a population proportion.

Some notes about conducting a hypothesis test:

• The null hypothesis $H_0$ is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
• The alternative hypothesis $H_a$ is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
• If the alternative hypothesis is a “less than”, then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
• If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
• If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
• Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of 0.05. Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
• The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.

Suppose the hypotheses for a hypothesis test are:

$\begin{eqnarray*} H_0: & & p=20 \% \\ H_a: & & p \gt 20\% \end{eqnarray*}$

Because the alternative hypothesis is a $\gt$, this is a right-tail test.  The p -value is the area in the right-tail of the distribution.

$\begin{eqnarray*} H_0: & & p=50 \% \\ H_a: & & p \neq 50\% \end{eqnarray*}$

Because the alternative hypothesis is a $\neq$, this is a two-tail test.  The p -value is the sum of the areas in the two tails of the distribution.  Each tail contains exactly half of the p -value.

$\begin{eqnarray*} H_0: & & p=10\% \\ H_a: & & p \lt 10\% \end{eqnarray*}$

Because the alternative hypothesis is a $\lt$, this is a left-tail test.  The p -value is the area in the left-tail of the distribution.

## Steps to Conduct a Hypothesis Test for a Population Proportion

• Write down the null and alternative hypotheses in terms of the population proportion $p$.  Include appropriate units with the values of the proportion.
• Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
• Collect the sample information for the test and identify the significance level.
• If $n \times p \geq 5$ and $n \times (1-p) \geq 5$, use the normal distribution with $\displaystyle{z=\frac{\hat{p}-p}{\sqrt{\frac{p \times (1-p)}{n}}}}$.
• If one of $n \times p \lt 5$ or $n \times (1-p) \lt 5$, use a binomial distribution.
• The results of the sample data are significant.  There is sufficient evidence to conclude that the null hypothesis $H_0$ is an incorrect belief and that the alternative hypothesis $H_a$ is most likely correct.
• The results of the sample data are not significant.  There is not sufficient evidence to conclude that the alternative hypothesis $H_a$ may be correct.
• Write down a concluding sentence specific to the context of the question.

## USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION PROPORTION

The p -value for a hypothesis test on a population proportion is the area in the tail(s) of distribution of the sample proportion.  If both $n \times p \geq 5$ and $n \times (1-p) \geq 5$, use the normal distribution to find the p -value.  If at least one of $n \times p \lt 5$ or $n \times (1-p) \lt 5$, use the binomial distribution to find the p -value.

If both $n \times p \geq 5$ and $n \times (1-p) \geq 5$:

• For x , enter the value for $\hat{p}$.
• For $\mu$ , enter the mean of the sample proportions $p$.  Note:  Because the test is run assuming the null hypothesis is true, the value for $p$ is the claim from the null hypothesis.
• For $\sigma$ , enter the standard error of the proportions $\displaystyle{\sqrt{\frac{p \times (1-p)}{n}}}$.
• For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.
• Use the appropriate technique with the norm.dist function to find the area in the left-tail or the area in the right-tail.

If at least one of $n \times p \lt 5$ or $n \times (1-p) \lt 5$:

• The p -value is found using the binomial distribution.
• For x , enter the number of successes.
• For n , enter the sample size.
• For p , enter the the value of the population proportion $p$ from the null hypothesis.
• For the logic operator , enter true .  Note:  Because we are calculating an at most probability, the logic operator is always true.
• For p , enter the the value of the population proportion $p$ in the null hypothesis.
• For the logic operator , enter true .  Note:  Because we are calculating an at least probability, the logic operator is always true.

Marketers believe that 92% of adults own a cell phone.  A cell phone manufacturer believes that number is actually lower.  In a sample of 200 adults, 87% own a cell phone.  At the 1% significance level, determine if the proportion of adults that own a cell phone is lower than the marketers’ claim.

Hypotheses:

$\begin{eqnarray*} H_0: & & p=92\% \mbox{ of adults own a cell phone} \\ H_a: & & p \lt 92\% \mbox{ of adults own a cell phone} \end{eqnarray*}$

From the question, we have $n=200$, $\hat{p}=0.87$, and $\alpha=0.01$.

To determine the distribution, we check $n \times p$ and $n \times (1-p)$.  For the value of $p$, we use the claim from the null hypothesis ($p=0.92$).

$\begin{eqnarray*} n \times p & = & 200 \times 0.92=184 \geq 5 \\ n \times (1-p) & = & 200 \times (1-0.92)=16 \geq 5\end{eqnarray*}$

Because both $n \times p \geq 5$ and $n \times (1-p) \geq 5$ we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a $\lt$, the p -value is the area in the left tail of the distribution.

 norm.dist 0.87 0.0046 0.92 sqrt(0.92*(1-0.92)/200) true

So the p -value$=0.0046$.

Conclusion:

Because p -value$=0.0046 \lt 0.01=\alpha$, we reject the null hypothesis in favour of the alternative hypothesis.  At the 1% significance level there is enough evidence to suggest that the proportion of adults who own a cell phone is lower than 92%.

• The null hypothesis $p=92\%$ is the claim that 92% of adults own a cell phone.
• The alternative hypothesis $p \lt 92\%$ is the claim that less than 92% of adults own a cell phone.
• The function is norm.dist because we are finding the area in the left tail of a normal distribution.
• Field 1 is the value of $\hat{p}$.
• Field 2 is the value of $p$ from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume $p=0.92$.
• Field 3 is the standard deviation for the sample proportions $\displaystyle{\sqrt{\frac{p \times (1-p)}{n}}}$.
• The p -value of 0.0046 tells us that under the assumption that 92% of adults own a cell phone (the null hypothesis), there is only a 0.46% chance that the proportion of adults who own a cell phone in a sample of 200 is 87% or less.  This is a small probability, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.  In other words, the proportion of adults who own a cell phone is most likely less than 92%.

A consumer group claims that the proportion of households that have at least three cell phones is 30%.  A cell phone company has reason to believe that the proportion of households with at least three cell phones is much higher.  Before they start a big advertising campaign based on the proportion of households that have at least three cell phones, they want to test their claim.  Their marketing people survey 150 households with the result that 54 of the households have at least three cell phones.  At the 1% significance level, determine if the proportion of households that have at least three cell phones is less than 30%.

$\begin{eqnarray*} H_0: & & p=30\% \mbox{ of household have at least 3 cell phones} \\ H_a: & & p \gt 30\% \mbox{ of household have at least 3 cell phones} \end{eqnarray*}$

From the question, we have $n=150$, $\displaystyle{\hat{p}=\frac{54}{150}=0.36}$, and $\alpha=0.01$.

To determine the distribution, we check $n \times p$ and $n \times (1-p)$.  For the value of $p$, we use the claim from the null hypothesis ($p=0.3$).

$\begin{eqnarray*} n \times p & = & 150 \times 0.3=45 \geq 5 \\ n \times (1-p) & = & 150 \times (1-0.3)=105 \geq 5\end{eqnarray*}$

Because both $n \times p \geq 5$ and $n \times (1-p) \geq 5$ we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a $\gt$, the p -value is the area in the right tail of the distribution.

 1-norm.dist 0.36 0.0544 0.3 sqrt(0.3*(1-0.3)/150) true

So the p -value$=0.0544$.

Because p -value$=0.0544 \gt 0.01=\alpha$, we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that the proportion of households with at least three cell phones is more than 30%.

• The null hypothesis $p=30\%$ is the claim that 30% of households have at least three cell phones.
• The alternative hypothesis $p \gt 30\%$ is the claim that more than 30% of households have at least three cell phones.
• The function is 1-norm.dist because we are finding the area in the right tail of a normal distribution.
• Field 2 is the value of $p$ from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume $p=0.3$.
• The p -value of 0.0544 tells us that under the assumption that 30% of households have at least three cell phones (the null hypothesis), there is a 5.44% chance that the proportion of households with at least three cell phones in a sample of 150 is 36% or more.  Compared to the 1% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the claim that 30% of households have at least three cell phones is most likely correct.

A teacher believes that 70% of students in the class will want to go on a field trip to the local zoo.  The students in the class believe the proportion is much higher and ask the teacher to verify her claim.  The teacher samples 50 students and 39 reply that they would want to go to the zoo.  At the 5% significance level, determine if the proportion of students who want to go on the field trip is higher than 70%.

$\begin{eqnarray*} H_0: & & p = 70\% \mbox{ of students want to go on the field trip} \\ H_a: & & p \gt 70\% \mbox{ of students want to go on the field trip} \end{eqnarray*}$

From the question, we have $n=50$, $\displaystyle{\hat{p}=\frac{39}{50}=0.78}$, and $\alpha=0.05$.

$\begin{eqnarray*} n \times p & = & 50 \times 0.7=35 \geq 5 \\ n \times (1-p) & = & 50 \times (1-0.7)=15 \geq 5\end{eqnarray*}$

Because both $n \times p \geq 5$ and $n \times (1-p) \geq 5$ we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a $\gt$, the p -value is the area in the right tail of the distribution.

 1-norm.dist 0.78 0.1085 0.7 sqrt(0.7*(1-0.7)/50) true

So the p -value$=0.1085$.

Because p -value$=0.1085 \gt 0.05=\alpha$, we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that the proportion of students who want to go on the field trip is higher than 70%.

• The null hypothesis $p=70\%$ is the claim that 70% of the students want to go on the field trip.
• The alternative hypothesis $p \gt 70\%$ is the claim that more than 70% of students want to go on the field trip.
• The p -value of 0.1085 tells us that under the assumption that 70% of students want to go on the field trip (the null hypothesis), there is a 10.85% chance that the proportion of students who want to go on the field trip in a sample of 50 students is 78% or more.  Compared to the 5% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the teacher’s claim that 70% of students want to go on the field trip is most likely correct.

Joan believes that 50% of first-time brides in the United States are younger than their grooms.  She performs a hypothesis test to determine if the percentage is the same or different from 50%.  Joan samples 100 first-time brides and 56 reply that they are younger than their grooms.  Use a 5% significance level.

$\begin{eqnarray*} H_0: & & p=50\% \mbox{ of first-time brides are younger than the groom} \\ H_a: & & p \neq 50\% \mbox{ of first-time brides are younger than the groom} \end{eqnarray*}$

From the question, we have $n=100$, $\displaystyle{\hat{p}=\frac{56}{100}=0.56}$, and $\alpha=0.05$.

To determine the distribution, we check $n \times p$ and $n \times (1-p)$.  For the value of $p$, we use the claim from the null hypothesis ($p=0.5$).

$\begin{eqnarray*} n \times p & = & 100 \times 0.5=50 \geq 5 \\ n \times (1-p) & = & 100 \times (1-0.5)=50 \geq 5\end{eqnarray*}$

Because both $n \times p \geq 5$ and $n \times (1-p) \geq 5$ we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a $\neq$, the p -value is the sum of area in the tails of the distribution.

Because there is only one sample, we only have information relating to one of the two tails, either the left or the right.  We need to know if the sample relates to the left or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, the sample proportion $\hat{p}=0.56$ is greater than the value of the population proportion in the null hypothesis $p=0.5$ ($\hat{p}=0.56>0.5=p$), so the sample information relates to the right-tail of the normal distribution.  This means that we will calculate out the area in the right tail using 1-norm.dist .  However, this is a two-tailed test where the p -value is the sum of the area in the two tails and the area in the right-tail is only one half of the p -value.  The area in the left tail equals the area in the right tail and the p -value is the sum of these two areas.

 1-norm.dist 0.56 0.1151 0.5 sqrt(0.5*(1-0.5)/100) true

So the area in the right tail is 0.1151 and  $\frac{1}{2}$( p -value)$=0.1151$.  This is also the area in the left tail, so

p -value$=0.1151+0.1151=0.2302$

Because p -value$=0.2302 \gt 0.05=\alpha$, we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that the proportion of first-time brides that are younger than the groom is different from 50%.

• The null hypothesis $p=50\%$ is the claim that the proportion of first-time brides that are younger than the groom is 50%.
• The alternative hypothesis $p \neq 50\%$ is the claim that the proportion of first-time brides that are younger than the groom is different from 50%.
• We use norm.dist($\hat{p}$,$p$,$\mbox{sqrt}(p*(1-p)/n)$,true) to find the area in the left tail.  The area in the right tail equals the area in the left tail, so we can find the p -value by adding the output from this function to itself.
• We use 1-norm.dist($\hat{p}$,$p$,$\mbox{sqrt}(p*(1-p)/n)$,true) to find the area in the right tail.  The area in the left tail equals the area in the right tail, so we can find the p -value by adding the output from this function to itself.
• The p -value of 0.2302  is a large probability compared to the 5% significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the claim that the proportion of first-time brides who are younger than the groom is most likely correct.

Watch this video: Hypothesis Testing for Proportions: z -test by ExcelIsFun [7:27]

An online retailer believes that 93% of the visitors to its website will make a purchase.   A researcher in the marketing department thinks the actual percent is lower than claimed.  The researcher examines a sample of 50 visits to the website and finds that 45 of the visits resulted in a purchase.  At the 1% significance level, determine if the proportion of visits to the website that result in a purchase is lower than claimed.

$\begin{eqnarray*} H_0: & & p=93\% \mbox{ of visitors make a purchase} \\ H_a: & & p \lt 93\% \mbox{ of visitors make a purchase} \end{eqnarray*}$

From the question, we have $n=50$, $x=45$, and $\alpha=0.01$.

To determine the distribution, we check $n \times p$ and $n \times (1-p)$.  For the value of $p$, we use the claim from the null hypothesis ($p=0.93$).

$\begin{eqnarray*} n \times p & = & 50 \times 0.93=46.5 \geq 5 \\ n \times (1-p) & = & 50 \times (1-0.93)=3.5 \lt 5\end{eqnarray*}$

Because $n \times (1-p) \lt 5$ we use a binomial distribution to calculate the p -value.  Because the alternative hypothesis is a $\lt$, the p -value is the probability of getting at most 45 successes in 50 trials.

 binom.dist 45 0.2710 50 0.93 true

So the p -value$=0.2710$.

Because p -value$=0.2710 \gt 0.01=\alpha$, we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that the proportion of visitors who make a purchase is lower than 93%.

• The null hypothesis $p=93\%$ is the claim that 93% of visitors to the website make a purchase.
• The alternative hypothesis $p \lt 93\%$ is the claim that less than 93% of visitors to the website make a purchase.
• The function is binom.dist because we are finding the probability of at most 45 successes.
• Field 1 is the number of successes $x$.
• Field 2 is the sample size $n$.
• Field 3 is the probability of success $p$.  This is the claim about the population proportion made in the null hypothesis, so that means we assume $p=0.93$.
• The p -value of 0.2710 tells us that under the assumption that 93% of visitors make a purchase (the null hypothesis), there is a 27.10% chance that the number of visitors in a sample of 50 who make a purchase is 45 or less.  This is a large probability compared to the significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the proportion of visitors to the website who make a purchase adults is most likely 93%.

A drug company claims that only 4% of people who take their new drug experience any side effects from the drug.  A researcher believes that the percent is higher than drug company’s claim.  The researcher takes a sample of 80 people who take the drug and finds that 10% of the people in the sample experience side effects from the drug.  At the 5% significance level, determine if the proportion of people who experience side effects from taking the drug is higher than claimed.

$\begin{eqnarray*} H_0: & & p=4\% \mbox{ of people experience side effects} \\ H_a: & & p \gt 4\% \mbox{ of people experience side effects} \end{eqnarray*}$

From the question, we have $n=80$, $\hat{p}=0.1$, and $\alpha=0.05$.

To determine the distribution, we check $n \times p$ and $n \times (1-p)$.  For the value of $p$, we use the claim from the null hypothesis ($p=0.04$).

$\begin{eqnarray*} n \times p & = & 80 \times 0.04=3.2 \lt 5\end{eqnarray*}$

Because $n \times p \lt 5$ we use a binomial distribution to calculate the p -value.  Because the alternative hypothesis is a $\gt$, the p -value is the probability of getting at least 8 successes in 80 trials.  (Note:  In the sample of size 80, 10% have the characteristic of interest, so this means that $80 \times 0.1=8$ people in the sample have the characteristic of interest.)

 1-binom.dist 7 0.0147 80 0.04 true

So the p -value$=0.0147$.

Because p -value$=0.0147 \lt 0.05=\alpha$, we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that the proportion of people who experience side effects from taking the drug is higher than 4%.

• The null hypothesis $p=4\%$ is the claim that 4% of the people experience side effects from taking the drug.
• The alternative hypothesis $p \gt 4\%$ is the claim that more than 4% of the people experience side effects from taking the drug.
• The function is 1-binom.dist because we are finding the probability of at least 8 successes.
• Field 1 is $x-1$ where $x$ is the number of successes.  In this case, we are using the compliment rule to change the probability of at least 8 successes into 1 minus the probability of at most 7 successes.
• Field 3 is the probability of success $p$.  This is the claim about the population proportion made in the null hypothesis, so that means we assume $p=0.04$.
• The p -value of 0.0147 tells us that under the assumption that 4% of people experience side effects (the null hypothesis), there is a 1.47% chance that the number of people in a sample of 80 who experience side effects is 8 or more.  This is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.  In other words, the proportion of people who experience side effects is most likely greater than 4%.

## Concept Review

The hypothesis test for a population proportion is a well-established process:

• Find the p -value (the area in the corresponding tail) for the test using the appropriate distribution (normal or binomial).
• Compare the p -value to the significance level and state the outcome of the test.

Teach yourself statistics

## Hypothesis Test for a Proportion

This lesson explains how to conduct a hypothesis test of a proportion, when the following conditions are met:

• The sampling method is simple random sampling .
• Each sample point can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
• The sample includes at least 10 successes and 10 failures.
• The population size is at least 20 times as big as the sample size.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

## State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

## Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the one-sample z-test to determine whether the hypothesized population proportion differs significantly from the observed sample proportion.

## Analyze Sample Data

Using sample data, find the test statistic and its associated P-Value.

σ = sqrt[ P * ( 1 - P ) / n ]

z = (p - P) / σ

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a z-score, use the Normal Distribution Calculator to assess the probability associated with the z-score. (See sample problems at the end of this lesson for examples of how this is done.)

## Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

In this section, two hypothesis testing examples illustrate how to conduct a hypothesis test of a proportion. The first problem involves a a two-tailed test; the second problem, a one-tailed test.

## Sample Size Calculator

As you probably noticed, the process of testing a hypothesis about a proportion can be complex. Stat Trek's Sample Size Calculator can do the same job quickly and easily. When you need to test a hypothesis, consider using the Sample Size Calculator. The calculator is free. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.

Problem 1: Two-Tailed Test

The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisified. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.80

Alternative hypothesis: P ≠ 0.80

• Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test .

σ = sqrt [(0.8 * 0.2) / 100]

σ = sqrt(0.0016) = 0.04

z = (p - P) / σ = (.73 - .80)/0.04 = -1.75

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the z-score is less than -1.75 or greater than 1.75. We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04. Since the standard normal distribution is symmetric with a mean of zero, we know that P(z > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.

• Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the sample included at least 10 successes and 10 failures, and the population size was at least 10 times the sample size.

Problem 2: One-Tailed Test Suppose the previous example is stated a little bit differently. Suppose the CEO claims that at least 80 percent of the company's 1,000,000 customers are very satisfied. Again, 100 customers are surveyed using simple random sampling. The result: 73 percent are very satisfied. Based on these results, should we accept or reject the CEO's hypothesis? Assume a significance level of 0.05.

Null hypothesis: P >= 0.80

Alternative hypothesis: P < 0.80

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100]

• Interpret results . Since the P-value (0.04) is less than the significance level (0.05), we cannot accept the null hypothesis.

## Module 8: Inference for One Proportion

Introduction to hypothesis test for a population proportion, what you’ll learn to do: conduct a hypothesis test for a population proportion..

## Contribute!

• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

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1. 8.8 Hypothesis Tests for a Population Proportion

The p -value for a hypothesis test on a population proportion is the area in the tail (s) of distribution of the sample proportion. If both n× p ≥ 5 n × p ≥ 5 and n ×(1− p) ≥ 5 n × ( 1 − p) ≥ 5, use the normal distribution to find the p -value. If at least one of n× p < 5 n × p < 5 or n×(1 −p) < 5 n × ( 1 − p) < 5, use ...

2. 8.4: Hypothesis Test Examples for Proportions

First, determine what type of test this is, set up the hypothesis test, find the p-value p -value, sketch the graph, and state your conclusion. Answer. Since the problem is about percentages, this is a test of single population proportions. H0: p = 0.85 H 0: p = 0.85. Ha: p ≠ 0.85 H a: p ≠ 0.85. p = 0.7554 p = 0.7554.

3. PDF Quarter 4 Module 14: of Hypothesis on Population Proportion

problems involving test of hypothesis on population proportion. The objective is for us to make a correct decision about the null hypothesis. It is whether we can confidently say that the change in our data is real, definite, and not attributed by chance. After going through this module, you are expected to: 1.

4. 3.4: Hypothesis Test for a Population Proportion

H1: p ≠ 0.75. Step 2) State the level of significance and the critical value. This is a two-sided question so alpha is divided by 2. Alpha is 0.05 so the critical values are ± Zα/2 = ± Z.025. Look on the negative side of the standard normal table, in the body of values for 0.025. The critical values are ± 1.96.

5. 8.15: Hypothesis Test for a Population Proportion (3 of 3)

Step 2: Collect the data. Since the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the sample proportion can be np ≥ 10 and n (1 − p) ≥ 10. Step 3: Assess the evidence. Determine the test statistic which is the z -score for the sample proportion.

6. PDF CHAPTER 8 Hypothesis Testing for Population Proportions

The notation used for the null hypothesis and alternative hypothesis Try to understand the difference between the significance level and the p-value. They are both probabilities! (See pp. 360 and 362) Know the formula for the one-proportion z-test statistic Section 8.2 Hypothesis Testing in Four Steps

7. PDF STAT 201 Chapter 9.1-9.2 Hypothesis Testing for Proportion

Hypothesis Test for Proportions: Step 5 (cont.) •For a right tailed test: 𝐻𝑎: > 0 We have rejection regions for 𝐻 are as follows •Note: all of the rejection region is in the right tail, where is much larger than 0 Confidence Reject (test stat) Reject (p-value) 0.90 Test-stat>1.282 P-value<.1 0.95 Test-stat>1.645 P-value<.05

8. Hypothesis Test for a Proportion

Test statistic. The test statistic is a z-score (z) defined by the following equation. z = (p - P) / σ. where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and σ is the standard deviation of the sampling distribution. P-value.

9. Hypothesis Test for a Population Proportion (1 of 3)

The research question helps us to form our hypotheses: H 0: p = 0.23 (No difference; the proportion of African Americans on juries in murder trials is the same as the proportion of African Americans in the population.) H a: p < 0.23 (The proportion of African Americans on juries in murder trials is less than the proportion of African Americans ...

10. Hypothesis Test for a Population Proportion (2 of 3)

Our sample proportion was 0.02 above the population proportion from the null hypothesis. In a sample of size 500, we would observe a sample proportion 0.02 or more away from 0.84 about 22% of the time by chance alone. Step 4: State a conclusion. Again we compare the P-value to the level of significance, α = 0.05.

11. Hypothesis Testing

This statistics video tutorial explains how to solve hypothesis testing problems with proportions. It explains how to calculate the sample proportion and th...

12. Hypothesis test comparing population proportions

Our alternative hypothesis is that there is a difference. Or that P1 does not equal P2. Or that P1 minus P2, the proportion of men voting minus the proportion of women voting, the true population proportions, do not equal 0. And we're going to do the hypothesis test with a significance level of 5%.

13. 8.7: Hypothesis Test of Single Population Proportion with Examples

Example 8.7.1 8.7. 1. Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms.

14. Hypothesis Test for a Population Proportion (3 of 3)

Step 2: Collect the data. Since the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the sample proportion can be np ≥ 10 and n (1 − p) ≥ 10. Step 3: Assess the evidence.

15. Solving Problems Involving Test of Hypothesis on Population Proportion

‼️fourth quarter‼️🟣 grade 11: solving problems involving test of hypothesis on population proportion‼️fourth playlists are already available‼️🟣 grade 11gen...

16. PDF Hypothesis Testing about a Population Proportion

Two-tailed test - Testing the claim of "not equal to". : p = a. o. α : p ≠ a. Specify the α-level of the test. The level of the test determines how rare an event must be in order to reject the null hypothesis. This is typically given to you in the statement of the problem. Typical values of α are .10, .05, and .01.

17. PDF Quarter 4 Module 14: Solving Pr of Hypothesis on Population Proportion

1. enumerate the steps in solving problems involving test of hypothesis on population proportion; and 2. solve problems involving test of hypothesis on the population proportion. What I Know Directions: Choose the best answer to the given questions or statements. Write the letter of your choice on a separate sheet of paper. 1.

18. 9.5: A Population Proportion

In a hypothesis test problem, you may see words such as "the level of significance is 1%." ... Just like with confidence intervals, we can make inferences using hypothesis tests involving proportions. ... A 0.01 level of significance means that $$\alpha = 0.01$$. This is a test of a single population proportion. $$H_{0}: p = 0.00078$$

19. Statistics and Probability Module: Solving Problems Involving Test of

Although problem solving has steps, someone may have his/her own way or techniques of solving a problem. Meanwhile, in statistical analysis, there are steps that need to be followed in solving problems involving test of hypothesis on population proportion. The objective is for us to make a correct decision about the null hypothesis.

20. Introduction to Hypothesis Test for a Population Proportion

When we have real world data on population proportions we will have to learn when a situation calls for testing a hypothesis about a population proportion, conduct a hypothesis test and state a conclusion in context. We will interpret the P-value as a conditional probability in the context of a hypothesis test. We will then distinguish the ...

21. Statistics and Probability Module: Formulating Appropriate Null and

2. formulate the appropriate null and alternative hypotheses concerning population proportions; and. 3. identify whether the given hypothesis test is a single-tailed or a two-tailed test. Statistics and Probability Quarter 4 Self-Learning Module: Formulating Appropriate Null and Alternative Hypotheses on a Population Proportion

22. Stat and Prob Q4 Mod14 Solving Problems Involving Test of Hypothesis on

Statistics and Probability - Grade 11 Alternative Delivery Mode Quarter 4 - Module 14: Solving Problems Involving Test of Hypothesis on Population Proportion First Edition, 2020. Republic Act 8293, Section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government ...

23. PDF Quarter 4 Module 8: Solving Problems Involving Test of Hypothesis on

raw cornstarch diet has a mean glucose level of 135 with a standard d. = 25 = 0.10Step 1: State the null and alternative hypothes. = 120 : ≠ 120Step 2: Determine the test statistic, then compute its value.= 24Since it is the population mean being tested, the population st. is unknown, a.