Quiz 6 Solution

Earlier this quarter we examined “risk ratings” from a random sample of 596 subjects. (In truth, the sample was not random, but let’s assume that it was.) One of the questions they were asked was to rate the risk level associated with living near a nuclear power plant. A rating of 0 means that there is no risk, and 100 is severe risk.

The average of risk of this sample was 77.86 and the standard deviation of the sample was 26.29.

Find an approximate 90% confidence interval for the mean risk in the population associated with living near a nuclear power plant. Show your work. (Some useful values are below.)

Some useful numbers: z(a) is defined to mean P(Z > z(a) ) = a where Z is a N(0,1) random variable.

z(.01) = 2.33; z(.025) = 1.96; z(.05) = 1.64; z(.075) = 1.44; z(.10) = 1.28;
z(.15) = 1.04


There was an inadvertant error in this problem statement. Really, the correct confidence interval is the one based on the t-distribution, because here the SD of the population is unknown, and instead we are estimating it with the SD of the sample (26.29). However, the correct value for the t-distribution was t(.05) with n-1 = 595 degrees of freedom. This is qt(.95, 595) = 1.647 which is quite close to z(0.05) = 1.64. So little is lost by using this normal approximation.

The CI is Xbar +- 1.64 (SD/sqrt(n))
77.86 +- 1.64 (26.29/sqrt(596)
77.86 +- 1.766
(76.09, 79.63)

Some of you did not give an interval, but instead just left it as 77.86 +- 1.766. I did not take points off, but will in the future. The reason is that (a) the problem asks for an interval and (b) the form (x,y) is easier to interpret. Often we look at confidence intervals and use them like hypothesis tests. I might wonder “Is 75 a legitimate value for the population mean?”. By glancing at the interval (76.09, 79.63) I can see in a flash that it is not (because it’s not in the interval.) But if I had to look at 77.86 +- 1.766 then I have to do a little arithmetic before I will know that 75 is not a likely value.