# Thread: I have given up on this question...help anyone

1. ## I have given up on this question...help anyone

An experiment is performed to measure the dependence between solubility of a chemical in water and the temperature of the water. The following results are obtained:

Temperature, x (oC) 0, 0, 0, 10, 10, 10, 20, 20, 20
Solubility, y (g/litre) 4.8, 4.5, 4.6, 10.3, 10.5, 10.3, 16.4, 17.8, 15.9
Temperature, x (oC) 30, 30, 30, 40, 40, 40, 50, 50, 50
Solubility, y (g/litre) 22.1, 23.6, 23.2, 28.9, 29, 28.1, 34.5, 35.2, 35.8

1)Fit a linear model to the data by estimating the parameters α and β.

2)A chemical process is to run at 16oC. Find a 95% confidence interval for the expected solubility of the chemical at this temperature.

2. Originally Posted by goaway716
An experiment is performed to measure the dependence between solubility of a chemical in water and the temperature of the water. The following results are obtained:

Temperature, x (oC) 0, 0, 0, 10, 10, 10, 20, 20, 20
Solubility, y (g/litre) 4.8, 4.5, 4.6, 10.3, 10.5, 10.3, 16.4, 17.8, 15.9
Temperature, x (oC) 30, 30, 30, 40, 40, 40, 50, 50, 50
Solubility, y (g/litre) 22.1, 23.6, 23.2, 28.9, 29, 28.1, 34.5, 35.2, 35.8

1)Fit a linear model to the data by estimating the parameters α and β.

2)A chemical process is to run at 16oC. Find a 95% confidence interval for the expected solubility of the chemical at this temperature.
1) You are asked to do a linear regression analysis, so do it.

CB

3. Originally Posted by goaway716
An experiment is performed to measure the dependence between solubility of a chemical in water and the temperature of the water. The following results are obtained:

Temperature, x (oC) 0, 0, 0, 10, 10, 10, 20, 20, 20
Solubility, y (g/litre) 4.8, 4.5, 4.6, 10.3, 10.5, 10.3, 16.4, 17.8, 15.9
Temperature, x (oC) 30, 30, 30, 40, 40, 40, 50, 50, 50
Solubility, y (g/litre) 22.1, 23.6, 23.2, 28.9, 29, 28.1, 34.5, 35.2, 35.8

1)Fit a linear model to the data by estimating the parameters α and β.

2)A chemical process is to run at 16oC. Find a 95% confidence interval for the expected solubility of the chemical at this temperature.
If you have ever used SAS(Statistical Analysis Software) or even MS-EXCEL(if you haven't heard of SAS), then this question should be a piece of cake!

Your linear model should look like this:

$y = \alpha + \beta x$ ; where y is the solubility and x is the temperature, α is constant and β is the parameter estimate for x(temperature). If β is positive, then with every unit increase in temperature, the solubility is predicted to be β units higher, holding other variables constant, and vice versa.

for confidence intervals, you need to have a nice look at your textbook!