Question

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?

Answer 1

Answer:

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

$df=n-1=25-1=24$

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 2

Answer: 1.711, we accept the null hypothesis.

Step-by-step explanation: this is a question under hypothesis testing for some sample mean.

Let H' be the null hypothesis and H1 be the alternative hypothesis.

H': u = 41,500

H1 : u > 41, 500 ( sample mean x = 43,780)

Where u = population mean = 41,500

s = sample standard deviation = 10,520

From the alternative hypothesis, we can see that the sample mean is greater than the population mean hence making the test upper tailed.

Our test statistics will be a t test and that's because sample size is lesser than 30 ( n = 25) and we are given our sample standard deviation.

We need our critical value and test statistics value to make conclusions.

To get the critical value, we make use of the degree of freedom ( n - 1 = 25 - 1 = 24) and the level of significance (5%).

We check the degree of freedom against the level of significance on a t distribution table.

By doing so we have our critical value as 1.711

Let us now get our test statistics.

The t score is calculated as

x - u / (s/√n)

= 43,780 - 41,500/ (10,520/√25)

= 2280/ (10,520/5)

= 2280/ 2104

= 1.084

By comparing our test statistics to the critical value, we can see that test statistics is lesser than the critical value ( 1.084 < 1.711), hence we accept the null hypothesis.