#This is a second example from Topic 16 # The problem is: # 1) we have a population with an unknown mean # and unknown standard deviation. # 2) we have a null hypothesis that the mean of the # population is 135.3 # 3) we have an alternative hypothesis that the mean # of the population is not equal to 135.3 # 4) we want to test the null hypothesis against the # alternative hypothesis at the 0.03 level of # significance. # 5) to do this we take a random sample of size 32 # 6) compute the sample mean and standard deviation; # sample mean = 138.8 sample standard deviation=9.2 # ----------------------- # 7) at this point we could use the critical value # approach and find the critical low value and critical # high values for the sample mean, the values that have # probability that a sample mean will be either less than # the critical low value or higher than the critical high # value = 0.03, The probabilities need to be split evenly. # 8) if the sample mean is less than the critical low # or higher than the critical high # value then we reject H0 in favor of H1, if not then # we do not have enough evidence to reject H0 in favor # of H1. # ------- alternatively ----- # 9) we could use the attained significance approach and # determine the probability of getting a sample mean # as extreme or more extreme than we just got. Note # that we need to specify extreme or more extreme, not # just lower or higher. # 10) if that probability is less than the level of # significance then we reject H0 in favor of H1, # if not the we do not have enough evidence to reject # H0 in favor of H1 ################################################ # for the critical value approach first # find the t value with 0.03/2 to its right # for 31 degrees of freedom high_t <- qt( 0.03/2, 31, lower.tail=FALSE ) high_t # by symmetry, the low t values will be the opposite low_t <- -high_t low_t # # our critical low will be 135.3 + low_t*9.2/sqrt( 32 ) # remember that low_t is negative # our critical high value will be 135.3 + high_t*9.2/sqrt( 32 ) # remember that high_t is positive # Because our sample mean, 138.8 is not less than the # critical low value and not higher than the # critical high we do not reject H0 in favor of H1 # or use the attained significance approach # First normalize the sample mean so that we can use # the pt() function. # standard_t <- (138.8-135.3)/(9.2/sqrt(32)) standard_t # this becomes our test statistic # find the probability of getting that value or more extreme. # because this is a high value we will use lower.tail=FALSE p_val <- pt( standard_t, 31, lower.tail=FALSE ) # That answer now needs to be doubled to account for # being this extreme on the other side (lower side in this case). p_val*2 # that answer, 0.03929236, is not less than our level of # significance, 0.03 so we do not reject H0 # in favor of H1 # # Or we could have just used our # hypoth_test_unknown function to compute both # approaches source("../hypo_unknown.R") # note the 0 to indicate that H1: mean != 135.3 hypoth_test_unknown( 135.3, 0, 0.03, 32, 138.8, 9.2)