Question

The fuel efficiency (in miles per gallon) of midsize minivans made by two auto companies is compared. Twenty test drivers are randomly divided into two groups of 10 people. One group of drivers took turn driving a car made by company A on the same route of a highway, and the other group of drivers did the same with a car made by company B. The company A car yielded an average gas mileage of 57.9 and a standard deviation of 3.4, whereas the company B car yielded an average gas mileage of 53.2 and a standard deviation of 2.8. Construct a 99% confidence interval for the difference in gas mileage between the two models. Assume normal populations with equal variance. Find the upper bound of the confidence interval (round off to first decimal place).

Answer 1

Answer:

$4.7-2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=1.775$  

$4.7+2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=7.625$  

So on this case the 95% confidence interval would be given by $1.775 \leq \mu_A -\mu_B \leq 7.625$  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X_A=57.9$ represent the sample mean A

$\bar X_B =53.2$ represent the sample mean B

n1=10 represent the sample A size  

n2=10 represent the sample B size  

$s_1 =3.4$ sample standard deviation for sample A

$s_2 =2.8$ sample standard deviation for sample B

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} s_p \sqrt{\frac{1}{n_A}+\frac{1}{n_B}}$ (1)

Where $s_p$ represent the standard deviation pooled given by:

$s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}$

$s_p =\sqrt{\frac{(10 -1)(3.4)^2 +(10-1)(2.8)^2}{10 +10 -2}}=3.114$

The point of estimate for $\mu_A -\mu_B$ is just given by:

$\bar X_A -\bar X_B =57.9-53.2=4.7$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n_A +n_B -2=10+10-2=18$  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.10$  

Confidence interval

Now we have everything in order to replace into formula (1):  

$4.7-2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=1.775$  

$4.7+2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=7.625$  

So on this case the 95% confidence interval would be given by $1.775 \leq \mu_A -\mu_B \leq 7.625$