Question

Are biomedical engineering salaries in Miami equal to those in Minnesota or is there a difference? A study was conducted looking at 40 staff engineering salaries from Miami and 45 staff engineering salaries in Minnesota:

Miami – sample size =40, average salary = 69,000, sample standard deviation = 1250

Minnesota – sample size = 45, average salary = 72,000, sample standard deviation = 1450.

At a 0.05 level of significance, can we conclude there is a difference in salaries based on the random sampled data? (assume that the standard deviations from the population although unknown are equal)

Use the six steps of hypothesis testing to make your decision.

Answer 1

Miami

Sample size = n1 = 40,

Average salary = X1 = 69,000

Sample standard deviation = s1 = 1250

Minnesota

Sample size = n2 = 45,

Average salary = X2 = 72,000,

Sample standard deviation = s = 1450.

In biomedical engineering, salaries in Miami equal to those in Minnesota is our null hypothesis and for alternate hypothesis we assume that they are unequal.

(a)Null and Alternate Hypothesis

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test.

(b)Assuming that the standard deviations from the population although unknown are equal. Given significance level is α=0.05, and the degrees of freedom is

df = 40+45-2 = 83

Critical t values

t lower = -1.989 using Excel formula =T.INV(0.025,83)

t upper = 1.989 using Excel formula =T.INV(0.975,83)

Rejection region will be R= -1.989 < t< 1.989

(c) Test statistics

(d) Since it is observed that t test = -10.153 < t lower= -1.989, it is then concluded that the null hypothesis is rejected.

(e) The p-value is p = 0 using excel formula =T.DIST(-10.153,83,TRUE) as value is ver very low we take thi as zero., and since p =0<0.05, it is concluded that the null hypothesis is rejected

(f)There is sufficient eveidence to reject the claim that in biomedical engineering, salaries in Miami is equal to those in Minnesota at significance level of 5%..