Question

An article in the ACI Materials Journal (Vol. 84, 1987, pp. 213-216) describes several experiments investigating the rodding of concrete to remove trapped air. A 3-inch x 6-inch cylinder was used, and the number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in the following table. Compressive Strength (psi) Rodding Level Observations 10 1530 1530 1440 15 1610 1650 1500 20 1560 1730 1530 25 1500 1490 1510 (a) Is there any difference in compressive strength due to the rodding level? Use a = 0.05. Calculate to 2 decimal places fo: Is there statistical evidence to say that rodding level affects compressive strength? (b) Find to 2 decimal places the P-value for the F-statistic computed in part (a). P-value = (c) Analyze the following residual plots to determine model adequacy. Does the assumption of normality seem reasonable? Does there appear to be an outlier in the data? Does the assumption of constant variance seem reasonable?

Answer 1

(A)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ_{4 or no difference in the rodding level.}

Ha: Not all means are equal or the rodding level of atleast two groups differ.

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df_1 = 3df1​=3 and df_2 = 3df2​=3, therefore, the rejection region for this F-test is R={F:F>Fc​=4.066}.

(3) Test Statistics

The following table is obtained:

	Group 1	Group 2	Group 3	Group 4
	1530	1610	1560	1500
	1530	1650	1730	1490
	1440	1500	1530	1510
Sum =	4500	4760	4820	4500
Average =	1500	1586.667	1606.667	1500
\sum_i X_{ij}^2 =∑i​Xij2​=	6755400	7564600	7767400	6750200
St. Dev. =	51.962	77.675	107.858	10
SS =	5400	12066.666666667	23266.666666667	200
n =	3	3	3	3

The total sample size is N = 12N=12. Therefore, the total degrees of freedom are:

dftotal​=12−1=11

Also, the between-groups degrees of freedom are dfbetween​=4−1=3, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=11−3=8

First, we need to compute the total sum of values and the grand mean. The following is obtained

∑​Xij​=4500+4760+4820+4500=18580

Also, the sum of squared values is

∑​Xij2​=6755400+7564600+7767400+6750200=28837600

Based on the above calculations, the total sum of squares is computed as follows

1 185802 SSIotal = ΣΧ ΣΧ, = 28837600 - 69566.667 N 12 1, 2.]

The within sum of squares is computed as shown in the calculation below:

SS within 200 = 40933.333 S Swithin groups 5400+12066.666666667+23266.666666667+

S Sbetween 28633.333 M Sbetween - 9544.444 dfbetween 3

40933.333 MS within S Swithin dfurithin 5116.667 8

F M Sbetween M Swithin 9544.444 5116.667 1.865

F(critical)=4.066

(4) Decision about the null hypothesis

Since it is observed that F=1.865≤Fc​=4.066, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.2138, and since p=0.2138≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the α=0.05 significance level or _{or no difference in the rodding level}

(B) P-value=0.2138

.

i hope it helps..

If you have any doubts please comment and please don't dislike.

PLEASE GIVE ME A LIKE. ITS VERY IMPORTANT FOR ME.