Question

A union leader in the Maritime Workers' Union believed that workers at the Port of Brisbane were receiving a lower weekly salary than workers at Port Kembla. To determine whether this claim was valid, they took samples of size 17 and 11 in Brisbane and Port Kembla, respectively, and found that the average and standard deviation of the weekly salaries were $914.11 and $49.38 respectively in Port Kembla, and $859.23 and $55.62 in Brisbane. Use Brisbane minus Port Kembla, and answer correct to two decimal places. 1. Determine a point estimate for the difference in the average weekly salary between the two ports. ? 2. Calculate the standard error for the difference between sample means assuming that the workers' salaries in both locations are normally distributed and have the same population variance. ? 3. Use Kaddstat to determine a 95% confidence interval for the difference between the mean weekly salaries in Brisbane and Port Kembla. lower limit ? upper limit ?

Answer 1

mean of sample 1,    x̅1=   914.1100
standard deviation of sample 1,   s1 =    49.3800
size of sample 1,    n1=   17
      
mean of sample 2,    x̅2=   859.23
standard deviation of sample 2,   s2 =    55.62
size of sample 2,    n2=   11

1)

estimate of difference in sample means =    x̅1 - x̅2 =    54.880

2)

pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    51.86891555

std error , SE =    Sp*√(1/n1+1/n2) =    20.0708

3)

Degree of freedom, DF=   n1+n2-2 =    26
t-critical value =    t α/2 =    2.0555

margin of error, E =    t*SE =    41.2562

estimate of difference in sample means =    x̅1 - x̅2 =    54.880

confidence interval is

Interval Lower Limit=   (x̅1-x̅2) - E =    13.6238
Interval Upper Limit=   (x̅1-x̅2) + E =    96.1362