An experiment was conducted to investigate the transition pressure of bismuth as a function of temperature. Listed below in Table 3.1 are the temperature (x) in degrees Centigrade and the difference in pressure (y) from 25,000 bars in hundreds of bars for 23 samples. (So if y = 3.66 then the pressure is 25, 000 + 3.66 × 100 = 25, 366 bars.)
Table 3.1 The temperature and pressure difference for bismuth transition.
You can assume the following calculations:
a) Produce an ANOVA table to test the 'lack of fit' and establish if it is reasonable to assume that the expected pressure is a linear function of the temperature.
b) Test the hypothesis that the slope is zero.
c) Obtain a point estimate and 95% confidence interval for the expected decrease in pressure due to an increase of 10 degrees centigrade.
A study was conducted to see how the amount of heat, which is generated when cement sets, is influenced by the composition of the cement. The response vari- able is the heat generated (y) in calories per gram. The explanatory variables are the percentages of cement by weight of the constituents tricalcium-silicate (x1) and tetracalcium-alumino-ferrite (x2) and the data for 13 cement samples are given in Table 3.2.
Table 3.2 Heat generated in calories per gram and percentages of cement by weight of the constituents tricalcium-silicate and tetracalcium-alumino-ferrite for 13 cement samples.
A multiple regression model was fitted to these data with heat generated as a linear function of x1 and x2 with an intercept term using MINITAB.
a) Obtain the ANOVA table for the model with predictors x1 and x2.
b) Which percentage of the variability of y is explained by the overall model?
c) Test the hypothesis that β1 = β2 = 0. What do you conclude?
d) Compare the model with two linear predictors to the one with a single predictor
x1 and decide which one is the best.
e) Using the Minitab listing, calculate βˆ0.
f) Test the hypothesis βi = 0 for each i = 0,1,2 individually. What do you conclude?
g) Calculate the 95% confidence intervals for β0, β1, β2.
Degree of freedom for regression is 2-1=1. MS(regression)=SS(regression)/1=429.3754
Degree of freedom for residuals is 22-1=21. MS(residual)=2.2146/21=0.105457
P value=FDIST(4071.569,1,21)= 1.56324E-25
Since P value is less than 0.05, we could reject null hypothesis and conclude that the expected pressure is a linear function of the temperature.
Meanwhile, based on the calculations above, we could have the ANOVA table below:
null hypothesis: b1=0
alternative hypothesis: b1≠0
this is a two tailed t test.
The degree of freedom is 23-2=21.
The critical t values are ...
The expert examines transition pressures of bismuth as a function of temperature.