Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
95% = (Mean1-Mean2) ± 1.96 SEdiff
Part 1)
Exposure to dust at work can lead to lung disease later in life. In a study of 115 drill and blast workers involved in tunnel construction the mean dust exposure was 18.0 mg-yr/m^3 with a standard deviation of 7.8 mg-yr/m^3 (sample 1). Another study of 220 outdoor concrete workers had a mean exposure of 6.5 mg-yr/m^3 with a standard deviation of 3.4 mg-yr/m^3 (sample 2). Test the claim that the exposures for these two types of workers are the same at a 5% level of significance.
Part 2)
Using the data from the previous problem, construct a 95% confidence interval about μ1 - μ2.
Any help will be greatly appreciated. Thanks for your time.
1 answer
1 answer