Question

A car company claims that their Super Spiffy Sedan averages 31 mpg. You randomly select 8 Super Spiffies from local car dealerships and test their gas mileage under similar conditions. You get the following MPG scores: MPG: 30 28 32 26 33 25 28 30 Does the actual gas mileage for these cars deviate significantly from 31 (alpha = .05)?

Answer 1

Answer:

$t=\frac{29-31}{\frac{2.777}{\sqrt{8}}}=-2.037$    

$p_v =2*P(t_{(7)}<-2.037)=0.081$  

If we compare the p value and the significance level given $\alpha=0.05$ 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 true mean is not significantly different from 31 at 1% of signficance

Step-by-step explanation:

Data given and notation  

Data: 30 28 32 26 33 25 28 30

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=29$ represent the mean height for the sample  

$s=2.777$ represent the sample standard deviation

$n=8$ sample size  

$\mu_o =31$ 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 conduct a hypothesis in order to check if the true mean is 31, the system of hypothesis would be:  

Null hypothesis:$\mu = 31$  

Alternative hypothesis:$\mu \neq 31$  

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:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (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:  

$t=\frac{29-31}{\frac{2.777}{\sqrt{8}}}=-2.037$    

P-value

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

$df=n-1=8-1=7$  

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

$p_v =2*P(t_{(7)}<-2.037)=0.081$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ 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 true mean is not significantly different from 31 at 1% of signficance