Question

In a large hospital, a nursing director selected a random sample of 30 registered nurses and found that the mean of their ages was 30.2. The population standard deviation for the ages is 5.6. She selected a random sample of 40 nursing assistants and found the mean of their ages was 31.7. The population standard deviation of the ages for the assistants is 4.3. Find the 99% confidence interval of the differences in the ages.

Answer 1

Solution:

Given that,

= 30.2

= 31.7

= 5.6

= 4.3

n₁ = 30

n₂ = 40

a) - is the point estimate  of the difference between two means .

point estimate =    - = 30.2 - 31.7 = -1.5

b) c = 99% = 0.99

= 2.576

Margin of error =  

= 2.576 * (5.6² /30) + ( 4.3² / 40 )

= 3.16

c) Confidence interval is

point estimate margin of error

-1.5 3.16

(-1.5- 3.16, -1.5+3.16)

(-4.66, 1.66)