# Solved:During 2006, 3.0% of all U.S. households were burglary victims. For a simple random sample of 300 households from a certain region

#### ByDr. Raju Chaudhari

Sep 26, 2020

During 2006, 3.0% of all U.S. households were burglary victims. For a simple random sample of 300 households from a certain region, suppose that 18 households were victimized by burglary during that year. Apply an appropriate hypothesis test and the 0.05 level of significance in determining whether the region should be considered as having a burglary problem greater than that for the nation as a whole.

### Solution

Given that $n = 300$, $X= 18$.

The sample proportion is

 \begin{aligned} \hat{p}&=\frac{X}{n}=\frac{18}{300}=0.06 \end{aligned}

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : p = 0.03$ against $H_1 : p > 0.03$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned}
which follows $N(0,1)$ distribution.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical values

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.64}$.

The rejection region (i.e. critical region) for the hypothesis testing problem is $\text{Z > 1.64}$.

#### Step 5 Computation

The test statistic is

 \begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.06-0.03}{\sqrt{\frac{0.03* (1-0.03)}{300}}}\\ & =3.046 \end{aligned}

#### Step 6 Decision (Traditional approach)

The test statistic is $Z =3.046$ which falls $inside$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision ($p$-value approach)

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($Z=3.046$). Thus the $p$-value = $P(Z<3.046)=0.0012$.

The p-value is $0.0012$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

We conclude that the region should be considered as having a burglary problem greater than that for the nation as a whole.