Question

1. A researcher suggests that male nurses earn more than female nurses. A survey of 16 male nurses and 20 female nurses reported the following data. Is there enough evidence to support the claim that male nurses earn more than female nurses? Use α=0.05. (4 points)

Female:   = $23,750

                S₁² = $250

                n₁ = 20

Male:        = $23,800

                S₂² = $300

                n₂ = 16

2. Create a 90% confidence interval for the difference between Male and Female salaries. ​

Answer 1

Question 1

Here, we have to use two sample t test for population means assuming equal population variances.

H₀: µ_female = µ_male versus H_a: µ_female < µ_male

α = 0.05

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given values, we have

S_p² = [(20 – 1)*250 + (16 – 1)*300]/(20 + 16 – 2)

S_p² = 74632.3529

t = (23750 – 23800) / sqrt[74632.3529*((1/20)+(1/16))]

t = -0.5457

df = n1 + n2 – 2 = 20 + 16 – 2 = 34

P-value = 0.2944

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is enough evidence to support the claim that male nurses earn more than female nurses.

Question 2

Confidence level = 90%

df = n1 + n2 – 2 = 20 + 16 – 2 = 34

Critical t value = 1.6909

(by using t-table)

Margin of error = E = t*sqrt[S_p²*((1/n1)+(1/n2))]

Margin of error = E = 1.6909* sqrt[74632.3529*((1/20)+(1/16))]

Margin of error = E = 154.9401

(X1bar – X2bar) = (23750 – 23800) = -50

Confidence interval = (X1bar – X2bar) ± E

Confidence interval = -50 ± 154.9401

Lower limit = -50 - 154.9401 = -204.9401

Upper limit = -50 + 154.9401 = 104.9401

Confidence interval = (-204.9401, 104.9401)