Question

A study was conducted by the Department of Zoology at the Virginia Tech to estimate the difference in the amount of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus was measured in milligrams per liter. Fifteen Specimens were collected from station 1, and 12 specimens were obtained from station 2. The 15 specimens from station 1 had and an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 specimens from station 2 had an average of content of 1.49 milligrams per liter and a standard deviation of 0.8 milligram per liter. Assume that the observations came from normally distributed populations with different variances. What formula should be used in order to find a 96% confidence interval for the difference in the true average orthophosphorus contests at those two stations? Are the assumptions satisfied for using the formula? # continued. Compute a 96% confidence interval for the difference in the true average orthophosphorus contests at those two stations? # continued. What conclusion can be drawn based on the resulted interval?

Answer 1

Solution

96% CONFIDENCE INTERVAL IS (0.511 , 4.819)

We need to construct the 96 % confidence interval for the difference between the population means  for the case that the population standard deviations are not known . The following information has been provided about each of the samples :

Based on the information provided , we assume that the population variances are unequal , so then the number of degrees of freedom is computed as follows :