# 6a.4.3 - Steps in Conducting a Hypothesis Test for $p$

6a.4.3 - Steps in Conducting a Hypothesis Test for $p$

## Six Steps for One-Sample Proportion Hypothesis Test

### Steps 1-3

Let's apply the general steps for hypothesis testing to the specific case of testing a one-sample proportion.

Step 1: Set up the hypotheses and check conditions.

$np_0\ge 5$ and $n(1−p_0)≥5$

One Proportion Z-test Hypotheses

Left-Tailed
$H_0\colon p=p_0$
$H_a\colon p<p_0$
Right-Tailed
$H_0\colon p=p_0$
$H_a\colon p>p_0$
Two-Tailed
$H_0\colon p=p_0$
$H_a\colon p\ne p_0$

Step 2: Decide on the level of significance $\boldsymbol{(\alpha)}$.

Step 3: Calculate the test statistic.

One Proportion Z-test: $z^*=\dfrac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

The first few steps (Step 1 - Step 3) are exactly the same as the rejection region or p-value approach. The next part will discuss steps 4 - 6 for both approaches.

## Rejection Region Approach

### Steps 4-6

Step 4: Find the appropriate critical values for the tests. Write down clearly the rejection region for the problem.

#### Two-Tailed Test

View the critical values and regions with an $\alpha=.05$.

Step 5: Make a decision about the null hypothesis.
Check to see if the value of the test statistic falls in the rejection region. If it does, then reject $H_0$ (and conclude $H_a$). If it does not fall in the rejection region, do not reject $H_0$.
Step 6: State an overall conclusion.

## P-Value Approach

### Steps 4-6

Step 4: Compute the appropriate p-value based on our alternative hypothesis.
Left-Tailed
$P(Z \le z^*)$
Right-Tailed
$P(Z\ge z^*)$
Two-Tailed
$2$ x $P(Z \ge |z^*|)$
Step 5: Make a decision about the null hypotheses.
If the p-value is less than the significance level, then reject the null hypothesis. If the p-value is greater than the significance level, fail to reject the null hypothesis.
Step 6: State an overall conclusion.
Note! Recall that the P-value is a probability of obtaining a value of the test statistic or a more extreme value of the test statistic assuming that the null hypothesis is true.

## Example 6-5: Penn State Students from Pennsylvania Referring back to example 6-4. Say we take a random sample of 500 Penn State students and find that 278 are from Pennsylvania. Can we conclude that the proportion is larger than 0.5 at a 5% level of significance?

Conduct the test using both the rejection region and p-value approach.

Step 1: Set up the hypotheses and check conditions.

Set up the hypotheses. Since the research hypothesis is to check whether the proportion is greater than 0.5 we set it up as a one (right)-tailed test:

$H_0\colon p=0.5$ vs $H_a\colon p>0.5$

Can we use the z-test statistic? The answer is yes since the hypothesized value $p_0$ is $0.5$ and we can check that: $np_0=500(0.5)=250 \ge 5$ and $n(1-p_0)=500(1-0.5)=250 \ge 5$

Step 2: Decide on the significance level, $\alpha$.

According to the question, $\alpha= 0.05$.

Step 3: Calculate the test statistic:

\begin{align} z^*&= \dfrac{0.556-0.5}{\sqrt{\frac{0.5(1-0.5)}{500}}}\\z^*&=2.504 \end{align}

#### Rejection Region Approach

Step 4: Find the appropriate critical values for the test using the z-table. Write down clearly the rejection region for the problem.

We can use the standard normal table to find the value of $Z_{0.05}$. From the table, $Z_{0.05}$ is found to be $1.645$ and thus the critical value is $1.645$. The rejection region for the right-tailed test is given by:

$z^*>1.645$

Step 5: Make a decision about the null hypothesis.

The test statistic or the observed Z-value is $2.504$. Since $z^*$ falls within the rejection region, we reject $H_0$.

Step 6: State an overall conclusion.

With a test statistic of $2.504$ and critical value of $1.645$ at a 5% level of significance, we have enough statistical evidence to reject the null hypothesis. We conclude that a majority of the students are from Pennsylvania.

#### P-Value Approach

Step 4: Compute the appropriate p-value based on our alternative hypothesis:
$\text{p-value}=P(Z\ge z^*)=P(Z \ge 2.504)=0.0062$

Step 5: Make a decision about the null hypothesis.

Since $\text{p-value} = 0.0062 \le 0.05$ (the $\alpha$ value), we reject the null hypothesis.

Step 6: State an overall conclusion.

With a test statistic of $2.504$ and p-value of $0.0062$, we reject the null hypothesis at a 5% level of significance. We conclude that a majority of the students are from Pennsylvania.

## Try it!

### Online Purchases

An e-commerce research company claims that 60% or more graduate students have bought merchandise online. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students shows that only 22 students have ever done so. Is there enough evidence to show that the true proportion is lower than 60%?

Conduct the test at 10% Type I error rate and use the p-value and rejection region approaches.

Step 1: Set up the hypotheses and check conditions.

Set up the hypotheses. Since the research hypothesis is to check whether the proportion is less than 0.6 we set it up as a one (left)-tailed test:

$H_0\colon p=0.6$ vs $H_a\colon p<0.6$

Can we use the z-test statistic? The answer is yes since the hypothesized value $p_0$ is 0.6 and we can check that: $np_0=80(0.6)=48 \ge 5$ and $n(1-p_0)=80(1-0.6)=32 \ge 5$

Step 2: Decide on the significance level, $\alpha$.

According to the question, $\alpha= 0.1$.

Step 3: Calculate the test statistic:

\begin{align} z^* &=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\\&=\frac{.275-0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}\\&=-5.93 \end{align}

#### Rejection Region Approach

Step 4: Find the appropriate critical values for the test using the z-table. Write down clearly the rejection region for the problem.

The critical value is the value of the standard normal where 10% fall below it. Using the standard normal table, we can see that the value is -1.28.

Step 5: Make a decision about the null hypothesis.

The rejection region is any $z^*$ such that $z^*<-1.28$ . Since our test statistic, -5.93, is inside the rejection region, we reject the null hypothesis.

Step 6: State an overall conclusion.

There is enough evidence in the data provided to suggest, at 10% level of significance, that the true proportion of students who made purchases online was less than 60%.

#### P-Value Approach

Step 4: Compute the appropriate p-value based on our alternative hypothesis:
$\text{p-value}=P(Z \le -5.93) = 0.0000000003$
Step 5: Make a decision about the null hypothesis.

Since our p-value is very small and less than our significance level of 10%, we reject the null hypothesis.

Step 6: State an overall conclusion.

There is enough evidence in the data provided to suggest, at 10% level of significance, that the true proportion of students who made purchases online was less than 60%.

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