iPad 14:53 * 55%. 33 total pt(s)) In a study to compare the visibility of paints

Question

iPad 14:53 * 55%. 33 total pt(s)) In a study to compare the visibility of paints used on highways, four different types of paint were tested at three different locations. It is to be expected that paint wear will differ depending on the quantity and type of traffic on the road, so Highway was used as a blocking factor. The following table gives the amount of visibility of paint after a set period of exposure to traffic and weather; visibility was measured on a standard scale, with larger values indicating greater visibility. t Type 1 Highway A 10 11.3 10.3 11.8 Highway B 16.2 19.2 19.2 Highway C7.6 9.9 9.8 93 10.139.77 11.23 Noting that two of the data points have not been shown, carry out calculations for the following questions using only a calculator and appropriate summary statistics. The overall data standard deviation is1.591. partial ANOVA table is given below. Compute the values needed to fill in the blanks. t Type Highway 82 otal Compute the value of the Paint Type degrees of freedom. 3 [1 pt(s)] ou are correct. r receipt no. is 155-2275Previous Tries Compute the value of the Highway degrees of freedom, 2 [1 pt(s)] ou are correct. r receipt no. is 155-614 Previous Tries Compute the value of the Error degrees of freedom. 5 [1 pt(s)) Tries 1/1 Previous Tries Compute the value of the Paint Type sum of squares. 764 [2 pt(s)) Submt Answer Tries 1/3 Previous Tries Compute the value of the Highway sum of squares. 7.6400 [1 pt(s)] are correct. r receipt no. is 150Prevlous Tries Compute the value of the Total sum of squares. [1 pt(s)] Sutoit Anser Tries 0/3 Compute the value of the Error sum of squares. (1 pt(s)1 Submit AnswerTries 0/3 Compute the value of the Paint Type mean sum of squares. [1 pt(s)) Submit Answer Tries 0/3

Explanation / Answer

here we have a table of data of paint type and highway, we run a ANOVA test and the output as below

Two-way ANOVA: y versus highway, paint

Source DF SS MS F P
highway 2 19.1117 9.55583 8.78 0.017
paint 3 9.3692 3.12306 2.87 0.126
Error 6 6.5283 1.08806
Total 11 35.0092

S = 1.043 R-Sq = 81.35% R-Sq(adj) = 65.81%

So we get

error df=6

paint SS=9.3692

TSS=35.0092

SSE=6.5283

MS(paint)=3.12306

MSE=1.08806

F(paint)=2.87

numerator df=6

denominator df=3

0.1<p-value<0.25

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