Question

Fifty new computer chips were tested for speed in a certain application. The average speed, in Mhz, for the new chips was 495.6 and the standard deviation was 19.4.  

1. Compute a 95% confidence interval for the average speed using the sample of 50 computer chips.
2. An older version of the chips had a speed of 481.2 Mhz. Can you conclude that the mean speed for the new chips is greater than the average speed of 481.2 Mhz of the old chips?
3. A second sample of 100 new chips were selected and tested. The observed average computer speed was 500.3 mhz with standard deviation equal to 20.3. Can you conclude that the results from the first study are consistent with the second larger study?

Answer 1

a)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   49          
't value='   tα/2=   2.010   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   19.4000   / √   50   =   2.7436
margin of error , E=t*SE =   2.0096   *   2.7436   =   5.5134
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    495.60   -   5.513419   =   490.0866
Interval Upper Limit = x̅ + E =    495.60   -   5.513419   =   501.1134
95%   confidence interval is (   490.087   < µ <   501.113   )

b)

yes, because both ends of interval are greater than 481.2

c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   99          
't value='   tα/2=   1.984   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   20.3000   / √   100   =   2.0300
margin of error , E=t*SE =   1.9842   *   2.0300   =   4.0280
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    500.30   -   4.027960   =   496.2720
Interval Upper Limit = x̅ + E =    500.30   -   4.027960   =   504.3280
95%   confidence interval is (   496.272   < µ <   504.328   )

yes, can conclude that the results from the first study are consistent with the second larger study