Question

According to a study by the American Pet Food Dealers association, 63 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial a random sample of 300 households showed 210 own pets. Does this data disagree with the Pet Food Dealers Association data? Use a .05 level of significance.

Answer 1

Answer

$z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511$  

$p_v =2*P(z>2.511)=0.012$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food association

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

X=210 represent the households with pets

$\hat p=\frac{210}{300}=0.7$ estimated proportion of households with pets

$p_o=0.63$ 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 true proportion is 0.63:  

Null hypothesis:$p=0.63$  

Alternative hypothesis:$p \neq 0.63$  

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:  

$z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511$  

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:  

$p_v =2*P(z>2.511)=0.012$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food association