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

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Answer:

[tex]z=\frac{0.395 -0.3}{\sqrt{\frac{0.3(1-0.3)}{200}}}=2.932[/tex]  

[tex]p_v =2*P(z>2.932)=0.0034[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] 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

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

[tex]p_o=0.3[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]p=0.3[/tex]  

Alternative hypothesis:[tex]p \neq 0.3[/tex]  

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

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

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

Calculate the statistic  

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

[tex]z=\frac{0.395 -0.3}{\sqrt{\frac{0.3(1-0.3)}{200}}}=2.932[/tex]  

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 [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>2.932)=0.0034[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] 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