"Using Technology to Perform a Two-Way ANOVA Test In Exercises...

Question

"Using Technology to Perform a Two-Way ANOVA Test In Exercises 15–18, use technology and the block design to perform a two-way ANOVA test. Use α=0.10. Interpret the results. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal.

[APPLET] Advertising A study was conducted in which a sample of 20 adults was asked to rate the effectiveness of advertisements. Each adult rated a radio or television advertisement that lasted 30 or 60 seconds. The block design shows these ratings (on a scale of 1 to 5, with 5 being extremely effective).

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Accepted Answer

Hello. In this video, we are told that a materials engineering firm is testing the braking strength of 2 different metal alloys treated with 2 different cooling methods. We are told that the cooling method is oil quench versus water quench. The alloy type is between alloys X and Y, and we have a total sample size of 20 with 5 samples per group. Based on the Enova test results, which of the following is the correct conclusion, and we want to assume a significance level of 0.05. So in order to interpret the significance of the interactions, alloy type, and the cooling method, what we want to do is you want to take a look at the corresponding p values for each of the rows and compare them to the given significance value. So let's go ahead and start with the interpretation of the interactions. Now here, the P value for the interactions is given to us as 0.2896. Now if we were to compare this to the given significance level, this p value is greater than the significance level of 0.05. So what this means is that we fail to reject a null hypothesis that there is no interaction between the two factors. So it does not appear that the breaking strings are affected by the interaction between the cooling method and the alloy type. So there doesn't appear to be an interaction effect. So that is going to be the interpretation of the interactions. Now we are going to repeat this process for the first row, which is going to be the cooling method. Now the p value for the cooling method is given to us as 0.0012. And if we were to compare this to the significance level, this is less than 0.05. So what this means is that the the p value is significantly less, so we are going to reject no hypothesis of no effect for the cooling method. And so what this means is that the braking strengths do appear to be affected by which cooling method is being used. So that's the interpretation of cooling method. And if we take a look at the alloy type, we are going to get the same result when comparing the P value to the alloy type. The P value for the alloy type is 0.0015, which is less than 0.05, and that tells us that the braking, the braking strength is affected by the alloy type as well. So what these results, what does this tell us? Well, the conclusion here is going to be that there is no significant interaction, but both the cooling method and the alloy type significantly affect the braking strength. Therefore, the correct solution to this problem is going to be option. Is going to be option B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Two-Way ANOVA

Block Design

Assumptions of ANOVA

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