Question

The Brinell hardness of ductile iron is known to be no more than 168. An engineer measured the Brinell hardness of 32 pieces of ductile iron that were subcritically annealed. The resulting data were: 170 167 174 179 179 187 179 183 156 163 156 187 156 167 156 174 183 179 174 179 170 159 187 165 190 175 177 160 178 149 169 171 a) Compute the Mean and Variance of the Sample above. b) The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 168. Therefore, he was interested in testing the hypotheses.(Hint: Use the standard level of significant).

Answer 1

Part A) :

R-code for finding mean and variance.

X=c(170,167,174,179,179,187,179,183,156,163,156,187,156,167,156,174,183,179,174,179,170,159,187,165,190,175,177,160,178,149,169,171)
n=length(X)
M=mean(X);M
V=var(X)*((n-1)/n);V
SD=sqrt(V);SD

Answers:

Mean : 171.8125

Variance : 113.4648

Standard Deviation : 10.65199

Part B) :

. Let Ho: The mean Brinell hardness of all such ductile - iron pieces is 168 in Let M= 168 Hii The mean Brinell hardness of all such ductile non pieces is greater than 168 iet 4x168 since sample size n= 32 x 30 we use L- test statistic il Z: X-M , Ñ = sample mean. olsun .o= sample standard deviation na sample size from &-command we comput X=171.8125 & r= Jvar = 10.6519 . Z. 171.8125-168 10.6519 1 32 72.02467 Let 2=0.05. i.e. at 54. level of significance Za = 1:64485 since Zy Za i.e. at 5 % level of significance we reject Ho re. null hypothesis & accept Hiie alternative hypothesis. i.e. The mean Brinell hardness of all such ductile iron pieces is greater than 168.