Question

We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental US in Exercise 5.19. The mean and standard deviation of the reported differences are 1.1 degrees and 4.9 degrees respectively.Calculate a 90% confidence interval for the average difference between the temperature measurements between 1968 and 2008.lower bound: ______ degrees(please round to two decimal places)upper bound: ______ degrees(please round to two decimal places)

Answer 1

Answer:

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$    

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$    

So on this case the 90% confidence interval would be given by (-0.30;2.50)  

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".

$\bar X=1.1$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=51 represent the sample size  

Solution to the problem

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:

$df=n-1=51-1=50$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,50)".And we see that $t_{\alpha/2}=2.02$

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

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$    

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$    

So on this case the 90% confidence interval would be given by (-0.30;2.50)