Question

Are nursing salaries in City A lower than those in City B? As reported by a newspaper, salary data show staff nurses in City A earn less than staff nurses in City B. Suppose that in a follow-up study of 40 staff nurses in City A and 50 staff nurses in City B you obtain the following results. Assume population variances are unknown and unequal.

City A	City B
n1 = 40	n2 = 50
x1 = $56,300	x2 = $59,400
s1 = $6,000	s2 = $7,000

(a) Formulate hypotheses so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in City A are significantly lower than for those in City B. Use

α = 0.05.

A. H₀: μ₁ − μ₂ ≤ 0

H_a: μ₁ − μ₂ > 0

B. H₀: μ₁ − μ₂ < 0

H_a: μ₁ − μ₂ ≠ 0

C. H₀: μ₁ − μ₂ ≠ 0

H_a: μ₁ − μ₂ = 0

D. H₀: μ₁ − μ₂ = 0

H_a: μ₁ − μ₂ < 0

E. H₀: μ₁ − μ₂ > 0

H_a: μ₁ − μ₂ = 0

(b) What is the value of the test statistic? (Round your answer to three decimal places.)

(c) What is the p-value? The degrees of freedom for this test are 87. (Round your answer to four decimal places.)

p-value =

(d)What is your conclusion?

A. Reject H₀. There is insufficient evidence to conclude that the salaries of staff nurses are lower in City A than in City B.

B. Do not Reject H₀. There is sufficient evidence to conclude that the salaries of staff nurses are lower in City A than in City B.     

C. Do not reject H₀. There is insufficient evidence to conclude that the salaries of staff nurses are lower in City A than in City B.

D. Reject H₀. There is sufficient evidence to conclude that the salaries of staff nurses are lower in City A than in City B.

Answer 1

a)

D. H0: μ1 − μ2 = 0

Ha: μ1 − μ2 < 0

b)

Sample #1   ---->   1          
mean of sample 1,    x̅1=   56300.00          
standard deviation of sample 1,   s1 =    6000          
size of sample 1,    n1=   40          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   59400.000          
standard deviation of sample 2,   s2 =    7000.00          
size of sample 2,    n2=   50          
                  
difference in sample means = x̅1-x̅2 =    56300.000   -   59400.0000   =   -3100.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1371.1309          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3100.0000   /   1371.1309   ) =   -2.2609

c) p-value =        0.01313   [ excel function: =T.DIST(t stat,df) ]

d)

D. Reject H0. There is sufficient evidence to conclude that the salaries of staff nurses are lower in City A than in City B.