Mathematical Methods for Physics and Engineering - Matematica.NET

31.7 HYPOTHESIS TESTINGP (t|H 0 )0.50.40.30.2N =3N =2N =10N =50.10−4 −3 −2 −1 0 1 2 3 4tFigure 31.11 Student’s t-distribution for various values of N. The brokencurve shows the standard Gaussian distribution for comparison.◮Ten independent sample values x i , i =1, 2,...,10, are drawn at random from a Gaussiandistribution with unknown mean µ and unknown standard deviation σ. The sample valuesare as follows:2.22 2.56 1.07 0.24 0.18 0.95 0.73 −0.79 2.09 1.81Test the null hypothesis H 0 : µ =0at the 10% significance level.For our null hypothesis, µ 0 = 0. Since for this sample ¯x =1.11, s =1.01 and N = 10, itfollows from (31.113) that¯xt =s/ √ N − 1 =3.33.The rejection region for t is given by (31.114) where t crit is such thatC N−1 (t crit )=1− α/2,and α is the required significance of the test. In our case α =0.1 andN = 10, and fromtable 31.3 we find t crit =1.83. Thus our rejection region for H 0 at the 10% significancelevel ist1.83.For our sample t =3.30 and so we can clearly reject the null hypothesis H 0 : µ =0atthislevel. ◭It is worth noting the connection between the t-test and the classical confidenceinterval on the mean µ. The central confidence interval on µ at the confidencelevel 1 − α is the set of values for which−t crit < ¯x − µs/ √ N − 1