Question

About 29% of all burglaries are through an open or unlocked door or window. A sample of 130 burglaries indicated that 87 were not via an open or unlocked door or window. At the 0.05 level of significance, can it be concluded that this differs from the stated proportion

Answer 1

Answer:

$z=\frac{0.331 -0.29}{\sqrt{\frac{0.29(1-0.29)}{130}}}=1.03$  

$p_v =2*P(z>1.03)=0.303$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of burglaries with via an open or unlocked door or window NOT differs from 0.29 or 29% .  

Step-by-step explanation:

Data given and notation

n=130 represent the random sample taken

X=130-87=43 represent the number of burglaries with via an open or unlocked door or window

$\hat p=\frac{43}{130}=0.331$ estimated proportion of burglaries with via an open or unlocked door or window

$p_o=0.29$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 29% of all burglaries are through an open or unlocked door or window.:  

Null hypothesis:$p=0.29$  

Alternative hypothesis:$p \neq 0.29$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.331 -0.29}{\sqrt{\frac{0.29(1-0.29)}{130}}}=1.03$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z>1.03)=0.303$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of burglaries with via an open or unlocked door or window NOT differs from 0.29 or 29% .  

Answer 2

Testing the hypothesis, it is found that since the p-value of the test is of 0.303 > 0.05, it cannot be concluded that it differs from the stated proportion.

At the null hypothesis, it is tested that the proportion does not differ from 0.29, that is:

$H_0: p = 0.29$

At the alternative hypothesis, it is tested if it differs, that is:

$H_1: p \neq 0.29$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are:

$p = 0.29, n = 130, \overline{p} = \frac{43}{130} = 0.3308$

The value of the test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.3308 - 0.29}{\sqrt{\frac{0.29(0.71)}{130}}}$

$z = 1.03$

We have a two-tailed test, as we are testing if the mean is different of a value, hence, the p-value is P(|z| > 1.03), which is 2 multiplied by the p-value of z = -1.03.

• Looking at the z-table, z = -1.03 has a p-value of 0.1515.

2 x 0.1515 = 0.3030

Since the p-value of the test is of 0.303 > 0.05, it cannot be concluded that it differs from the stated proportion.

A similar problem is given at https://brainly.com/question/24166849