A study was made to determine if humidity conditions have an...

A study was made to determine if humidity conditions have an effect on the force required to pull apart pieces of glued plastic. Three types of plastic are tested using 4 different levels of humidity. The results, in kilograms, are as follows: (a) Assuming a model I experiment, perform an analysis of variance and test the hypothesis of no interaction between humidity and plastic type at the 0.05 level of significance. (b) Using only plastics $A$ and $B$ and the value of $s^{2}$ from part (a), once again test for the presence of interaction at the 0.05 level of significance. (c) Use a single-degree-of-freedom comparison and the value of $s^{2}$ from part (a) to compare, at the 0.05 level of significance, the force required at $30 \%$ humidity versus $50 \%, 70 \%,$ and $90 \%$ humidity. (d) Using only plastic $C$ and the value of $s^{2}$ from part (a), repeat part (c).

A study was made to determine if humidity conditions have an effect on the force required to pull apart pieces of glued plastic. Three types of plastic are tested using 4 different levels of humidity. The results, in kilograms, are as follows:
(a) Assuming a model I experiment, perform an analysis of variance and test the hypothesis of no interaction between humidity and plastic type at the 0.05 level of significance.
(b) Using only plastics $A$ and $B$ and the value of $s^{2}$ from part (a), once again test for the presence of interaction at the 0.05 level of significance.
(c) Use a single-degree-of-freedom comparison and the value of $s^{2}$ from part (a) to compare, at the 0.05 level of significance, the force required at $30 \%$ humidity versus $50 \%, 70 \%,$ and $90 \%$ humidity.
(d) Using only plastic $C$ and the value of $s^{2}$ from part
(a), repeat part (c).

Key Concepts

Single-Degree-of-Freedom Comparisons

Single-degree-of-freedom comparisons are planned contrasts used to test specific hypotheses about differences between treatment means based on one degree of freedom. These tests focus on targeted comparisons, for instance, contrasting one group level with the average of others, and provide a more precise inference about differences when total degrees-of-freedom are limited in a factorial design.

Hypothesis Testing in ANOVA

Hypothesis testing in ANOVA typically involves formulating a null hypothesis stating that all group means are equal (or that there is no interaction), and an alternative hypothesis that at least one group mean differs. The test statistic, usually an F-ratio, is compared against critical values corresponding to a specified significance level to decide whether to reject the null hypothesis.

Fixed Effects Model (Model I Experiment)

A fixed effects model assumes that the levels of the factor under investigation are the only levels of interest, and inferences are restricted to these levels. In a Model I experiment, all factors are treated as fixed, which means that results and interpretations are specific to the inspected levels, and randomness is attributed to experimental error.

Interaction Effects

The interaction effect in a factorial experiment indicates whether the effect of one factor on the response variable changes across the levels of another factor. Testing for interaction involves determining whether the combination of different factor levels produces a different outcome than would be expected from their independent effects, and it is typically evaluated using an ANOVA component.

Factorial Experimental Design

A factorial experimental design involves studying the effects of two or more factors simultaneously, with each factor having multiple levels. This design facilitates the examination of not only the main effects of each factor but also the possible interactions between factors, which helps in understanding how the effects of one factor may depend on the level of another.

Analysis of Variance (ANOVA)

ANOVA is a statistical technique used to compare multiple group means simultaneously to determine if at least one group significantly differs from the others. It decomposes the total variability in the data into variability due to treatment effects and error, allowing researchers to assess the overall significance of the effects in the experiment.

Transcript

00:03 This question is 14 .3.

00:08 Says a study was made to determine if humidity conditions have an effect on the force required to pull up purposes of plastic.

00:17 Three types of plastics are tested using four different levels of humidity.

00:28 The results in kilograms is follows.

00:33 The results in kilograms are as follows with the plastic type with a, b, and c.

00:40 Then you meet it with 50%, i mean 30%, 50%, 70%, 70%, and the results are given in kilograms there.

00:50 All right, then he says, assuming a model one experiment, perform an analysis of variance and test the hypothesis of no interaction between humidity, and plastic type at the 0 .05 level of significance there so i did the anova there this is my my presentation i did the anova the new the anova these are the f values that i came up with right they computed f values right 16 .9 for the plastic type 11 .36 for the humidity and 5 .29 for the interactions and they give me this p -value 0 .003, 0 .008, for humidity 0 .007 for interaction, 0 .003 for the plastic.

01:50 So with this figures from the anova, but conclusion, but we're saying the interaction is very much significant day.

01:59 Let's see the interaction there, the direction figure that we get, so it's very much significant there.

02:05 You can see it's a very small number then it's a small number for the p then b says using only plastics a and b and the values of the variance from part a once again test for the presence of interaction at the 0 .05 level of significance there so it be i said right this sum of squares for ab with only plastic type a and b the sum of squares it becomes 24 .89 there with three degrees of freedom there.

02:56 So hence the f value becomes 24 .89 divided by 3 over 3 all divided by 4 .2163 all all divided by right.

03:11 4 comma right okay these are the values that we get from like the 4 comma 2163 is the error value there so we are dividing to get the decrease of freedom is 3 so we divide by 3 then in further divide by 4 comma on 3 4 2 1 3 which is the the figure for the error day this mean square for the error there okay, so if you do that, what we get, we get an f value of 1 .97 there with the p value and the p value comes to 0 .1727...

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