Question

It is claimed that automobiles are driven on average more than 18 comma 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 18 comma 390 kilometers and a standard deviation of 4100 ​kilometers? Use a​ P-value in your conclusion.

Answer 1

Answer:

$t=\frac{18390-18000}{\frac{4100}{\sqrt{18}}}=0.404$  

$p_v =P(t_{(17)}>0.404)=0.346$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 18000 at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

$\bar X=18390$ represent the sample mean

$s=4100$ represent the sample standard deviation  

$n=18$ sample size  

$\mu_o =18000$ 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 check if the mean is higher than 18000, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 18000$  

Alternative hypothesis:$\mu > 18000$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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:  

$t=\frac{18390-18000}{\frac{4100}{\sqrt{18}}}=0.404$  

P-value  

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=18-1=17$  

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

$p_v =P(t_{(17)}>0.404)=0.346$  

Conclusion  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 18000 at 5% of significance.