Question

A publisher reports that 46%46% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 220220 found that 50%50% of the readers owned a particular make of car. Is there sufficient evidence at the 0.050.05 level to support the executive's claim?

Answer 1

`z=\frac{0.5 -0.46}{\sqrt{(0.46(1-0.46))/(220)}}=1.19`

`p_v =P(z>1.19)=0.136`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of the readers owned a particular make of caris not significantly higher than 0.46 so the claim is not statsitically supported

Explanation:

Data given and notation

n=220 represent the random sample taken

`\hat p=0.5` estimated proportion of the readers owned a particular make of car

`p_o=0.46` 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 the true proportion is higher than 0.5.:

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p > 0.5`

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{0.5 -0.46}{\sqrt{(0.46(1-0.46))/(220)}}=1.19`

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 right tailed test the p value would be:

`p_v =P(z>1.19)=0.136`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of the readers owned a particular make of caris not significantly higher than 0.46 so the claim is not statsitically supported