Mathematics, 14.04.2020 19:47 itsseand23

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?

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Answer from: quece233

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

p_v =P(z1.19)=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

Step-by-step 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:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:

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

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(z1.19)=0.136

Answer from: Quest

answer: c ?

step-by-step explanation:

Answer from: Quest

i think the length is 12 and the width is 17

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