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.

Accepted Solution

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)