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 168 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 225 hours. Find a 95% confidence interval for the difference between the mean lifetimes of the two types of transistors.

(115.2642, 222.7358).

Step-by-step explanation:

Given data:

type A: n_1=60, xbar_1=1827, s_1=168

type B: n_2=180, xbar_2=1658, s_2=225

n_1 = sample size 1, n_2= sample size 2

xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.

a=0.05, |Z(0.025)|=1.96 (from the  standard normal table)

So 95% CI is

(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]

=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)

= (115.2642, 222.7358).

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76

-4

step-by-step explanation:

do 6-10=

if u need it to be negative.