Answer:
a) = x1`-x2`/ Sp √ ( 1/n₁ + 1/n₂)
b) it is assumed that the population standard deviations are known (which is sometimes met and sometimes not known).
c) The Confidence interval = [ 19131.77, 36868.23]
Step-by-step explanation:
Part a:
The 99% confidence interval for the difference between the average salaries of male and female neurologists is given by
(x1`-x2`) ±t∝/2(υ)* √ ( Sp²/n₁ + Sp²/n₂)
= x1`-x2`/ Sp √ ( 1/n₁ + 1/n₂)
where Sp ²= s₁² ( n1-1) + s₂²(n2-1) /(n1-1) + (n2-1)
Part B:
1) the sample should be normally distributed
2) it is assumed that the population standard deviations are known (which is sometimes met and sometimes not known).
Part C:
Calculations
1) H0: μ₁=μ₂ against the claim Ha: μ₁ ≠μ₂
2)The test statistic is
t= x1`-x2`/ √ ( Sp²/n₁ + Sp²/n₂)
where
Sp ²= s₁² ( n1-1) + s₂²(n2-1) /(n1-1) + (n2-1)
= 15,000²(17) + 22,000²(22) / 37
Sp ²= 103378640
3) D.f = n1+n2-2 =21+18-2 = 37
4) for 2 tailed test the critical value of t is obtained by
t≥ t∝/2 ( d.f)=t≥ t0.005 (37)= ±2.7154= ± 2.72
5) Putting the values
(x1`-x2`) ±t∝/2(υ)* √ ( sp²/n₁ + sp²/n₂)
28000 ± (2.72 * 3265.89)
=[ 19131.77, 36868.23].
The Confidence interval = [ 19131.77, 36868.23]
