An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-temperature performance. A sample of 60 transistors of type A were tested and were found to have a mean lifetime of 1827 hours and a standard deviation of 164 hours. A sample of 180 transistors of type B were tested and were found to have a mean lifetime of 1658 hours and a standard deviation of 234 hours. Let μX represent the population mean for transistors of type A and μY represent the population mean for transistors of type B. Find a 95% confidence interval for the difference μX−μY . Round the answers to three decimal places.

95% confidence interval for the difference in population mean between the two transistors is a lower limit of 141.443 hours and an upper limit of 196.557 hours.

Step-by-step explanation:

Confidence interval is given as difference in mean +/- error margin (E)

Sample 1

mean = 1827 hours

sd = 164 hours

n1 = 60

Sample 2

mean = 1658 hours

sd = 234 hours

n2 = 180

difference in mean = 1827 - 1658 = 169 hours

pooled sd = [(60-1)164 + (180-1)234] ÷ (60+180-2) = 51,562 ÷ 238 = 216.65 hours

degree of freedom = n1+n2-2 = 60+180-2 = 238

confidence level = 95%

Critical value (t) corresponding 238 degrees of freedom and 95% confidence level is 1.97048.

E = t×pooled sd/√(n1+n2) = 1.97048×216.65/√240 = 27.557 hours

Lower limit of difference in mean = 169 - 27.557 = 141.443 hours

Upper limit of difference in mean = 169 + 27.557 = 196.557 hours

95% confidence interval is (141.443, 196.557)

(23*2)-12=x     x=34

step-by-step explanation:

12 cups   is the answer

step-by-step explanation: