A study was conducted to estimate the difference in the amount of phosphate measured at two different stations on the River. phosphorus is measured in milligrams per liter (mg/L). Fifteen samples were collected from Station 1 and 12 samples were collected from Station 2. The 15 samples from Station 1 had an average phosphorus content of 3.84 mg/L and a standard deviation of 3.07 mg/L, while the 12 samples from Station 2 had an average phosphorus content of 1.49 mg/L and a standard deviation of 0.80 mg/L. Determine a 90% confidence interval for the difference in the true average.

Question

anonymous · Answer

contents at these two stations, assuming that the observations came from normal populations with different variances.

anonymous · Answer

http://www.stat.yale.edu/Courses/1997-98/101/meancomp.htm Read the second part, Tests of Significance for Two Unknown Means and Unknown Standard Deviations. In general your confidence interval will be in the form [ Sample Statistic + or - Critical value * Standard deviation of statistic ]. You will need to use a t-test since you have sample standard deviation. Find the t-critical value for a 90% confidence interval. The site has the equation for the standard deviation

anonymous · Answer

How do i find t (v,alpha/2)? It's for a confidence interval test. I know the answer to a similar problem was t(16,0.025)=2.12

But I need to find t(16.3, 0.05)

anonymous · Answer

Where does the 2.12 come from?