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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.

User Emil Lerch
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1 Answer

4 votes

Answer:


t=(18390-18000)/((4100)/(√(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.

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=(\bar X-\mu_o)/((s)/(√(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=(18390-18000)/((4100)/(√(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.

User Avinash Sonee
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8.7k points