Question

company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26 miles per gallon. To see if the goal is being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 25.02 mpg and a standard deviation of 4.83 mpg. Is this strong evidence that they have failed to attain their fuel economy goal

Answer 1

Answer:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$    

$p_v =P(t_{(49)}<-1.435)=0.0788$  

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 fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

$\bar X=25.02$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size  

$\mu_o =26$ 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 at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 26$  

Alternative hypothesis:$\mu < 26$  

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:  

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$    

P-value

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

$df=n-1=50-1=49$  

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

$p_v =P(t_{(49)}<-1.435)=0.0788$  

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 fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Answer 2

Using the t-distribution, it is found that since the test statistic is more than the critical value for the left-tailed test, this is not strong evidence that they have failed to attain their fuel economy goal.

At the null hypothesis, it is tested if the fleet average is of at least 26 miles per gallon, that is:

$H_0: \mu \geq 26$

At the alternative hypothesis, it is tested if it is less, that is:

$H_1: \mu < 26$

We have the standard deviation for the sample, hence, the t-distribution is used to solve this question.

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

For this problem, the values of the parameters are: $\overline{X} = 25.02, \mu = 26, s = 4.83, n = 50$.

Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{25.02 - 26}{\frac{4.83}{\sqrt{50}}}$

$t = -1.43$

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05 and 50 - 1 = 49 df is of $t^{\ast} = -1.677$

Since the test statistic is more than the critical value for the left-tailed test, this is not strong evidence that they have failed to attain their fuel economy goal.

To learn more about the t-distribution, you can take a look at https://brainly.com/question/13873630