Question

10 Next 5 6 7 3 4 Previous 1 Question 2 of 12 (1 point) In a large hospital, a nursing director selected a random sample of 42 registered nurses and found that the mean of their ages was 32.0. The population standard deviation for the ages is 4.3. She selected a random sample of 31 nursing assistants and found the mean of their ages was 30.2. The population standard deviation of the ages for the assistants is 5.9. Find the 95 % confidence interval of the differences in the ages. Use, for the mean age of the nursing assistants. Round the answers to one decimal place. rn

Answer 1

n1 = 42

$\bar{x1}$ = 32

$\sigma1$ = 4.3

n2 = 31

$\bar{x2}$ = 30.2

= 5.9

a2

We have to find 95% confidence interval for difference in population mean.

Formula is

l - x2)- Eul-p2<(x1- r2)E

Here E is the margin of error.

E Zc + n1 n2

Zc = 1.96

( From z table)

E 1.96 0.4402381.122903 2.4505

l -x2) (32 30.2) 1.8

So confidence interval is ( 1.8 - 2.4505 , 1.8 + 2.4505) => ( - 0.7 , 4.3)

0.7l-2< 4.3

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