Mathematics, 02.04.2020 01:30 nihadsalim10

Suppose that the average number of miles that a car is driven in the first three years is 41,500. To determine whether leased cars are driven more than "average", 25 leased cars were sampled and the average number of miles on the leased cars was 43,780 with a standard deviation of 10,520. To determine whether the average number of miles on leased cars is statistically significantly greater than the number of miles on other cars (using the 5% level of significance), what is the critical value? (please round your answer to 3 decimal places)

what can you conclude concerning the null hypothesis?

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Answer from: aidan2073

$t=\frac{43780-41500}{\frac{10520}{\sqrt{25}}}=1.083$

df=n-1=25-1=24

We can calculate the critical value using the t distribution with 24 degrees of freedom and looking for a quantile that accumulates 0.05 of the area on the right and we got:

$t_{crit}= 1.711$

If we compare the statistic calculated and the critical value we see that $t_{calc}$ so then we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=43780$ represent the sample mean

s=10520 represent the sample standard deviation

n=25 sample size

$\mu_o =41500$ represent the value that we want to test

$\alpha=0.05$ 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 apply a one lower tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \leq 41500$

Alternative hypothesis : $\mu 41500$

Compute the test statistic

The statistic for this case 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:

$t=\frac{43780-41500}{\frac{10520}{\sqrt{25}}}=1.083$

Give the appropriate conclusion for the test

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

df=n-1=25-1=24

We can calculate the critical value using the t distribution with 24 degrees of freedom and looking for a quantile that accumulates 0.05 of the area on the right and we got:

$t_{crit}= 1.711$

Conclusion

If we compare the statistic calculated and the critical value we see that $t_{calc}$ so then we don't have enough evidence to reject the null hypothesis at 5% of significance.

Answer from: Quest

umumumumum

im not sure

Answer from: Quest

answer: -0.5 would be the most likley outcome

step-by-step explanation:

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Suppose that the average number of miles that a car is driven in the first three years is 41,500. To...