Can 5/12 be simlified

Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre

choli [55]

Answer:

$z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717$

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

$\hat p=\frac{46}{115}=0.4$ estimated proportion of people that have traded in their old car

$p_o=0.48$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.9

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 proportion is less than 0.48.:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

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:

$z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717$

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 $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.

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3 years ago

I need help with this

son4ous [18]

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