Question

It is claimed that automobiles are driven on average more than 21 comma 000 kilometers per year. To test this​ claim, 120 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 21 comma 900 kilometers and a standard deviation of 4100 ​kilometers? Use a​ P-value in your conclusion.

Answer 1

Answer:

$t=\frac{21900-21000}{\frac{4100}{\sqrt{120}}}=2.405$  

$p_v =P(t_{(119)}>2.405)=0.0089$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean its significant higher than 21000 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=21900$ represent the mean production for the sample  

$s=4100$ represent the sample standard deviation for the sample  

$n=36$ sample size  

$\mu_o =21000$ 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 more than 21000, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 21000$  

Alternative hypothesis:$\mu > 21000$  

If we analyze the size for the sample is > 30 but 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:  

$t=\frac{21900-21000}{\frac{4100}{\sqrt{120}}}=2.405$  

P-value  

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

$df=n-1=120-1=119$  

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

$p_v =P(t_{(119)}>2.405)=0.0089$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean its significant higher than 21000 at 5% of signficance.