Question

Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the previous 6 months, in a sample of 115 new-car buyers, 46 have traded in their old car. To determine (at the 10% level of significance) whether the proportion of new-car buyers that trade in their old car has statistically significantly decreased, what can you conclude concerning the null hypothesis?

Answer 1

Answer:

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

$p_v =P(z<-1.717)=0.0429$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ 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 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:  

$p_v =P(z<-1.717)=0.0429$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ 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 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.