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

`z=\frac{0.395 -0.3}{\sqrt{(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<\alpha` 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

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=(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 \\eq 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{(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{(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<\alpha` 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