Question

Each car in a sample of seven cars was tested for​ nitrogen-oxide emissions​ (in grams per​ mile), and the following results were obtained. 0.07​, 0.11​, 0.15​, 0.13​, 0.12​, 0.07​, 0.13 a. Assuming that this sample is representative of cars in​ use, construct a​ 95% confidence interval estimate of the mean amount of​ nitrogen-oxide emissions for all cars.

Answer 1

Answer:

The 95% confidence interval would be given by (0.0825;0.1395)  

Step-by-step explanation:

1) 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".

Data: 0.07​, 0.11​, 0.15​, 0.13​, 0.12​, 0.07​, 0.13

We can calculate the sample mean and deviation with the following formulas:

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

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

$\bar X=0.111$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=0.0308 represent the sample standard deviation

n=7 represent the sample size  

2) Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

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

$df=n-1=7-1=6$

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(0.025,6)".And we see that $t_{\alpha/2}=2.45$

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

$0.111-2.45\frac{0.0308}{\sqrt{7}}=0.0825$    

$0.111+2.45\frac{0.0308}{\sqrt{7}}=0.1395$

So on this case the 95% confidence interval would be given by (0.0825;0.1395)