Question

A car company claims that its cars achieve an average gas mileage of at least 26 miles per gallon. A random sample of eight cars form this company have an average gas mileage of 25.5 miles per gallon and a standard deviation of 1 mile per gallon. At α=0.06, can the company’s claim be supported, assuming this is a normally distributed data set?

Answer 1

`t=(25.5-26)/((1)/(√(8)))=-1.414`

The degrees of freedom are given by:

`df=n-1=8-1=7`

The p value for this case is given by:

`p_v =P(t_((7))<-1.414)=0.100`

Since the p value is higher than the significance level of 0.06 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 25.5 and then the claim makes sense

Explanation:

Information given

`\bar X=25.5` represent the sample mean

`s=1` represent the sample standard deviation

`n=8` sample size

`\mu_o =26` represent the value to verify

`\alpha=0.06` represent the significance level

t would represent the statistic (variable of interest)

`p_v` represent the p value

Hypothesis to est

We want to test if the true mean is at least 26 mpg, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 25.5`

Alternative hypothesis:
`\mu < 25.5`

The statistic for this case is given by;

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(25.5-26)/((1)/(√(8)))=-1.414`

The degrees of freedom are given by:

`df=n-1=8-1=7`

The p value for this case is given by:

`p_v =P(t_((7))<-1.414)=0.100`

Since the p value is higher than the significance level of 0.06 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 25.5 and then the claim makes sense