Question

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.

Paint Type	1	2	3	4
Highway A	10	11.3	10.3	11.8
Highway B	6.2	9.2	9.2	?
Highway C	7.6	9.9	9.8	?
	7.93	10.13	9.77	10.57

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 is 1.46. A partial ANOVA table is given below. Compute the values needed to fill in the blanks.

Source	DF	SS	MS	F
Paint Type
Highway			4.99
Error
Total

Answer 1

degree of freedom for paint type = 4-1=3 degree of freedom for Highway = 3-1 = 2 degree of freedom for total a total number of observation - 12-1 11. for error = 11-3-2 degree of freedom ss for parint - 347.93 +10,13 + 9.77 +10.57)/12 - ğ = 9.6 ss paint - & lgp gr -37(7.93.-9.6)2+(10:13 -9.612+ (9.77-9.61"+(10.57-9.6137 - 3[2.7889 +1.06 + 0.34+1.94] - 36.1289] : = 18.3867 SS Highway - 4.99x2 = 9.98 SST · multiply overall data variance sy by (n-1) = 11 = 2.59 x 11 = 28.49 MS for pent = 18.3867/3 = 6.1289 SSE = 28.49 – 18.3867-9.98 = 0.1233 MSE = 0.1233/6 -0.02055

ms paint = = 298.2433 6.1289 0.02055 point = MSE Fibahay - = 242.8224 Flighway a morigh ms highway MSE 4.99 0.02055 Source DE SS . MS I F Paint Type 3 18.39 | 6.13 | 298.24 9.98 | 4.99 1 242.82. Error 9. 121 0.021 Total 1 28.49