Question

It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly selected automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 23,500 kilometers and a standard deviation of 3900 kilometers? Use a P-value in your conclusion.

Answer 1

Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$      

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

$\bar X=23500$ represent the sample mean  

$s=3900$ represent the sample standard deviation      

$n=100$ sample size      

$\mu_o =2000$ represent the value that we want to test    

$\alpha$ represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:$\mu \leq 2000$      

Alternative hypothesis:$\mu > 2000$      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

$df=n-1=100-1=99$  

Since is a one-side upper test the p value would be:      

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.