Since we want to be 95% confident that our interval contains the true value of #mu#, we set #alpha = 0.05#. Then the #100(1-alpha)%# confidence interval for #mu# is found by the formula:

#bar x +- t_(alpha//2, n–1)sigma/sqrtn#

#=137 " ft"+-t_(0.025,31)(7.0" ft")/(sqrt 32)#

The #t#-value comes from Student's #t#-distribution that has #n-1# degrees of freedom; the area under the #t#-curve to the right of that #t#-value is 0.025. The value for #t_(alpha//2, n–1)# is usually found by lookup in a #t#-table, but for values of #n>30#, we know #t_(alpha//2, n–1) ~~ z_(alpha//2)#, so we can look up this #z#-value instead. (As #n# grows larger, #t# approaches the standard normal distribution #Z#.)

So our 95% confidence interval for #mu# becomes

#137 " ft"+-z_(0.025)(7.0" ft")/(sqrt 32)#

#~~137 " ft"+-(1.96)(1.237" ft")#

#~~137 " ft"+-2.425 " ft"#.