Question

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a GPS navigation system if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the GPS navigation system. You were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 at a 1% level of significance

Answer 1

Answer:

$z=\frac{0.395 -0.3}{\sqrt{\frac{0.3(1-0.3)}{200}}}=2.932$  

$p_v =2*P(z>2.932)=0.0034$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of people that they would have purchased the GPS navigation system is significantly different from 0.3

Step-by-step explanation:

Data given and notation

n=200 represent the random sample taken

X=79 represent the  number of people that they would have purchased the GPS navigation system

$\hat p=\frac{79}{200}=0.395$ estimated proportion of people that they would have purchased the GPS navigation system

$p_o=0.3$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 the true proportion is 0.3 or no.:  

Null hypothesis:$p=0.3$  

Alternative hypothesis:$p \neq 0.3$  

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:  

$z=\frac{0.395 -0.3}{\sqrt{\frac{0.3(1-0.3)}{200}}}=2.932$  

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 $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z>2.932)=0.0034$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of people that they would have purchased the GPS navigation system is significantly different from 0.3