Question

1. A manufacturer of steel wants to test the effect of the method of manufacture on the tensile strength of a particular type of steel. Four different methods have been tested and the data shown in Table 1. (Use Minitab)

1. Develop the ANOVA table and test the hypothesis that methods affect the strength of the cement. Use a = 0.05
2. Use the Tukey’s method with a =0.05 to make comparisons between pairs of means.
3. Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of the normality assumption?
4. Prepare a scatter plot of the results to aid the interpretation of the results of this experiment
5. Find a 95 percent confidence interval on the mean tensile strength of each method. Also find a 95 percent confidence interval on the difference in means for techniques 1 and 3. Does this aid in interpreting the results of the experiment?

                

                     Table 1

                                         Method                      Tensile Strength

                                              1                 6.5       7.6       7.5        6.0

                                              2                 9.8       9.7       8.6        8.9

                                              3                 7.7       6.2       6.9        7.0

4    9.0       8.8       8.5        9.5

Answer 1

from the given data of information

a manufacturer of steel wants to test the effect of the method of manufacturer on the tensile strength of a particular type of steel

by using four different methods have been tested

a.

by ANOVA table

source DF_ SS_ MS_ F _P

factor_3_19.108_6.369_16.87_0.000

error_12_4.530_0.378

b.

results of tuckys method

(1) subtracted form

lower center upper

2 1.0597 2.3500 3.6403

3- 1.2403 0.0500 1.3403

4 0.7597 2.0500 3.3403

(2) subtracted form

lower center upper

3 -3.5903 -2.3000 -1.0097

4 -1.5903 -0.3000 0.9903

(3) subtracted form

lower center upper

4.07097 2.0000 3.2903

we can observe from above results

the means of 1 and 3 are equal and

the mean of 2 and 4 are equal

c.

the normal probability plot is shown below

the normal probability plot of residual is approximately straight line

assumption is valid in this case

d.

scatter plot for method 1,2,3,4 is shown

e.

variable = 1

mean = 6.9

std deviation = 0.779

95% CI = 5.661, 8.139

variable =2

mean = 9.25

std deviation = 0.592

95% CI = 8.309, 10.191

variable = 3

mean = 6.95

std deviation = 0.614

95% CI = 5.973, 7.927

variable =4

mean = 8.95

std deviation = 0.42

95% CI = 8.281, 9.619

the difference in means of 1 and 3 is 95% CI for difference is

-1.325, 1.225

yes, this aid in interpretating the results is

confidence interval for difference in 1 and 3 include 0. from this we can say that their means are same as we have already interpretade in part b using tukeys test