The length of life in hours of a sample of 40 100watt light
The length of life (in hours) of a sample of 40 100-watt light bulbs manufactured in a Sylvania plant and a sample of 40 100-watt light bulbs manufactured in a Phillips plant are contained in tab Bulb Life in the Excel file. In order to analyze the difference in bulb life from these two manufacturers, determine whether the variances are equal and then run the appropriate t-test to test the hypothesis that the mean bulb life of Sylvania bulbs is different than the mean bulb life of Phillips bulbs at the .05 level of significance. State your conclusion and the basis for your conclusion.
| Sylvania | Phillips |
| 684 | 819 |
| 831 | 907 |
| 859 | 952 |
| 893 | 994 |
| 922 | 1016 |
| 939 | 1038 |
| 972 | 1096 |
| 1016 | 1153 |
| 697 | 836 |
| 835 | 912 |
| 860 | 959 |
| 899 | 1004 |
| 924 | 1018 |
| 943 | 1072 |
| 977 | 1100 |
| 1041 | 1154 |
| 720 | 888 |
| 848 | 918 |
| 868 | 962 |
| 905 | 1005 |
| 926 | 1020 |
| 946 | 1077 |
| 984 | 1113 |
| 1052 | 1174 |
| 773 | 897 |
| 852 | 942 |
| 870 | 986 |
| 909 | 1007 |
| 926 | 1022 |
| 954 | 1077 |
| 1005 | 1113 |
| 1080 | 1188 |
| 821 | 903 |
| 852 | 943 |
| 876 | 992 |
| 911 | 1015 |
| 938 | 1034 |
| 971 | 1082 |
| 1014 | 1116 |
| 1093 | 1230 |
Solution
Set Up Hypothesis
Null,Ho: u1 = u2
Alternative,mean bulb life of Sylvania bulbs is differentthan the mean bulb
life of Phillips - H1: u1 != u2
Test Statistic
X(Mean)=909.65
Standard Deviation(s.d1)=94.3052 ; Number(n1)=40
Y(Mean)=1018.35
Standard Deviation(s.d2)=96.9014; Number(n2)=40
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (39*8893.471 + 39*9389.881) / (80- 2 )
S^2 = 9141.676
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=909.65-1018.35/Sqrt((9141.676( 1 /40+ 1/40 ))
to=-108.7/21.38
to=-5.084
| to | =5.084
Critical Value
The Value of |t | with (n1+n2-2) i.e 78 d.f is 1.991
We got |to| = 5.084 & | t | = 1.991
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -5.0843 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

