The tensile strength of Portland cement is being studied. Four different mixing techniques can be used economically. A completely randomized experiment was conducted and the following data were recorded:
# Setting up data frame for problem 3.7
MixingExp <- data.frame(
Tstrength = c(3129,3000,2865,2890,3200,3300,2975,3150,
2800,2900,2985,3050,2600,2700,2600,2765),
MixMethod = as.factor((rep(1:4,each = 4))),
Index = as.factor(rep(1:4,4)))Test the hypothesis that mixing techniques affect the strength of the cement. Use \(\alpha = 0.05\).
\(H_0: u_1 = u_2 = u_3 = u_4\)
\(H_a\): at least one of the \(u_i\)’s differ from the other \(u_i\)’s.
We will be using the AOV (Analysis of Variance) function in R to analyze our data.
MixAnalysis <- (aov(Tstrength ~ MixMethod, data = MixingExp))
MixTable <- summary(MixAnalysis)| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| MixMethod | 3 | 489740.2 | 163246.73 | 12.72811 | 0.0004887 |
| Residuals | 12 | 153908.2 | 12825.69 |
The AOV resulted in \(p = 0.00049\) which is \(< 0.05\). Because the p-value is much smaller than alpha, we reject the null hypothesis and accept the alternative that mixing techniques do affect the strength of the cement.
A product developer is investigating the tensile strength of a new synthetic fiber that will be used to make cloth for men’s shirts. Strength is usually affected by the percentage of cotton used in the blend of materials for the fiber. The engineer conducts a completely randomized experiment with five levels of cotton and replicates the experiment five times. The data are shown in the following table.
# Setting up data frame for problem 3.10
TensileStr <- data.frame(observations = c(7,7,15,11,9,12,17,12,18,18,14,19,19,
18,18,19,25,22,19,23,7,10,11,15,11),
PercentCotton = as.factor((rep(seq(15,35,5),each = 5))),
Index = as.factor(rep(1:5,5)))Is there evidence to support the claim that cotton content affects the mean tensile strength? Use \(\alpha = 0.05\).
\(H_0: u_1 = u_2 = u_3 = u_4 = u_5\)
\(H_a\): at least one of the \(u_i\)’s differ from the other \(u_i\)’s.
Like before, we will be using the AOV (Analysis of Variance) function in R to analyze our data.
CottonAnalysis <- (aov(observations ~ PercentCotton, data = TensileStr))
CottonTable <- summary(CottonAnalysis)| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| PercentCotton | 4 | 475.76 | 118.94 | 14.75682 | 9.1e-06 |
| Residuals | 20 | 161.20 | 8.06 |
The AOV resulted in \(p = 9.1e-06\) which is \(< 0.05\). Because the p-value is much smaller than alpha, we reject the null hypothesis and accept the alternative that cotton content does affect the mean tensile strength.
An article in the ACI Materials Journal (Vol. 84, 1987, pp.213-216) describes several experiments investigating the rodding of concrete to remove entrapped 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:
# Setting up data frame for problem 3.20
RoddingExp <- data.frame(CompStr = c(1530,1530,1440,1610,1650,1500,
1560,1730,1530,1500,1490,1510),
RoddingLevel = as.factor((rep(seq(10,25,5),each = 3))),
Index = as.factor(rep(1:3,4)))Is there any difference in compressive strength due to the rodding level? Use \(\alpha = 0.05\).
\(H_0: u_1 = u_2 = u_3 = u_4\)
\(H_a\): at least one of the \(u_i\)’s differ from the other \(u_i\)’s.
Like before, we will be using the AOV (Analysis of Variance) function in R to analyze our data.
RoddingAnalysis <- (aov(CompStr ~ RoddingLevel, data = RoddingExp))
RoddingTable <- summary(RoddingAnalysis)| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| RoddingLevel | 3 | 28633.33 | 9544.444 | 1.865364 | 0.2137815 |
| Residuals | 8 | 40933.33 | 5116.667 |
The AOV resulted in \(p = 0.2138\) which is \(> 0.05\). Because the p-value is larger than alpha, we fail to reject the null hypothesis and conclude that there is not a difference in compressive strength due to the rodding level.
Find the P-value for the F statistic in part a.
The P-value and F statistic were calculated in Part a using the AOV function in R. The p-value was calculated as \(0.2138\) and the F statistic was calculated as \(1.8654\).