Question

3) The 14 daily maximum temperatures recorded for December 18th at the Weather Station of the University at City A between 1998-2011 are approximately normally distributed with a sample mean of -1.03°C and a sample standard deviation of 3.55°C. Assume that these data are independent random samples from a normal distribution (which ignores any temporal correlations or climatic trends). Compute a) 98% confidence interval for both the population mean b) and 98% confidence interval for the population standard deviation of December 18th daily maximum temperatures at this location. (10 marks)

Answer 1

(a) CONFIDENCE INTERVAL FOR POPULATION MEAN given that, sample mean, x=-1.03 ; standard deviation, s =3.55 ; sample size, n =14 level of significance, alpha = 0.02 from standard normal table, two tailed value of It alpha/2 with n-1 = 13 d.f is 2.65 we use CI = x + t a/2 * (sd/Sqrt(n) where, x = mean sd = standard deviation a = 1 - (confidence level/100) ta/2 = t-table value CI = confidence interval confidence interval = [ -1.03 Et a/2 ( 3.55/ Sart (14) ] = [-1.03-(2.65 * 0.949), -1.03+(2.65 * 0.949) ] = [-3.544 , 1.484 ]

(b) CONFIDENCE INTERVAL FOR STANDARD DEVIATION ci = (n-1) s^2 / 4^2 right < 0^2 < (n-1) s^2 / 4^2 left where, S = standard deviation y^2 right = (1 - confidence level)/2 y^2 left = 1 - y^2 right n = sample size since alpha =0.02 Y^2 right = (1 - confidence level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01 Y^2 left = 1 - y^2 right = 1 -0.01 = 0.99 the two critical values y^2 left, y^2 right at 13 df are 27.6882, 4.107 s.d( s )=3.55 sample size(n)=14 confidence interval for o^2= [ 13 * 12.6025/27.6882 < 0^2 < 13 * 12.6025/4.107 ] = [ 163.8325/27.6882 < 0^2 < 163.8325/4.1069 ] [ 5.9171 < 0^2 < 39.892 ] and confidence interval for o = sqrt(lower) < 0 < sqrt(upper) = [sqrt (5.9171) <o< sqrt(39.892), ) = [ 2.4325 <<6.316 ]