Question

Two different hardening processes, (1) saltwater quenching and (2) oil quenching, are used on samples of a particular type of metal allow. The results are shown here. Assume that hardness is normally distributed. Saltwater Quench 112 116 105 124 109 118 107 122 Oil Quench 115 103 120 108 113 97 119 107 105 a. Test the hypothesis that the mean hardness for the saltwater quenching process equals the b. Assuming that the variances σ' and σ, are equal, construct a 95% confidence interval on Does the assumption made carlier mean hardness for the oil quenching process. Use α=0.05 and assume equal variances. the difference in mean hardness. Construct a 95% confidence interval on the ratio of equal variances seem reasonable? c.

Answer 1

a) 9 X1 112 116 105 124 109 118 107 122 b) X2 115 103 120 108 113 97 119 107 105 s1: 6.6165 x1 bar 114.4444 (x1 bar - x2 bar) (-/+) E 4.7777(-/ 2.120 3.3752) n2 9 4.7777 (-/+) 7.1554 -2.38 and 11.93 s2: 7.6649 x2 bar 109.6667 Significancelevel:0.050 The 95% confidence interval for the mean differences is (-2.38 and 11.93) 0.05 | Mean = | 114.4444 | 109.6667 SD-I 6.6165 7.6649 c) difference of x1 bar and x2 bar is 4.7777 Sx1-x2((n1-1)s12+(n2-1)s22)/(n1+n2-2)(1/1+1/n2) F (1-ou2) 0.226 F (al2)4.433 95% interval 0.168 to 3.303 Sx1-x23.3752 For a total a- 0.050 With df=n1 + n2-2-16 With a/2- 0.025 critical t-score2.120 Here we find that the value of 1 falls in the above interval, this support the assumptions of the equal variances t(x1 bar -x2 bar) Sx1-x2- 1.416 tl1.416 P-value 0.1759 0.05 Ho cannot be rejected There is sufficient evidence to conclude that the mean hardness of the saltwater quenching process equals the mean hardness for the oil quenching process at 5% level of significance