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.

Answer:[tex]1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30[/tex]    [tex]1.1+2.02\frac{4.9}{\sqrt{50}}=2.50[/tex]    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". [tex]\bar X=1.1[/tex] represent the sample mean for the sample  [tex]\mu[/tex] 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: [tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1) In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by: [tex]df=n-1=51-1=50[/tex] Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], 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 [tex]t_{\alpha/2}=2.02[/tex] Now we have everything in order to replace into formula (1): [tex]1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30[/tex]    [tex]1.1+2.02\frac{4.9}{\sqrt{50}}=2.50[/tex]    So on this case the 90% confidence interval would be given by (-0.30;2.50)