Question

Test the claim that the proportion of people who own dogs is less than 32%. A random sample of 1000 people found that 28% owned dogs. Do the sample data provide convincing evidence to support the claim

Answer 1

Answer:

$z=\frac{0.28 -0.32}{\sqrt{\frac{0.32(1-0.32)}{1000}}}=-2.71$  

$p_v =P(z<-2.71)=0.00334$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v<\alpha$ 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 people who owned dogs is significantly less than 0.32.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

$\hat p=0.28$ estimated proportion of people who owned dogs

$p_o=0.32$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that proportion of people who own dogs is less than 0.32 or 32%"

Null hypothesis:$p\geq 0.32$  

Alternative hypothesis:$p < 0.32$  

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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.28 -0.32}{\sqrt{\frac{0.32(1-0.32)}{1000}}}=-2.71$  

4) 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 next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z<-2.71)=0.00334$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v<\alpha$ 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 people who owned dogs is significantly less than 0.32.