91影视

The table of data is

The fitted values are obtained by formula

\({\hat x_{ij}} = {\bar x_{i.}} + {\bar x_{.j}} - {\bar x_{..}}\)

The residuals are obtained by formula

\({x_{ij}} - {\hat x_{ij}} = {x_{ij}} - {\bar x_{i.}} - {\bar x_{.j}} + {\bar x_{..}}\)

There are total of 20 data points, to avoid every single calculation it will be shown how to compute 3 random residuals

\({x_{21}} - {\hat x_{21}} = {x_{21}} - {\bar x_{2.}} - {\bar x_{.1}} + {\bar x_{..}} = 722 - 843.5 - 755.8 + 866.25 = - 11.05\)

\({x_{32}} - {\hat x_{32}} = {x_{32}} - {\bar x_{3.}} - {\bar x_{.2}} + {\bar x_{..}} = 802 - 839 - 841.6 + 866.25 = - 12.35\)

\({x_{33}} - {\hat x_{33}} = {x_{33}} - {\bar x_{3.}} - {\bar x_{.3}} + {\bar x_{..}} = 880 - 839 - 908.2 + 866.25 = - 0.95\)

The corresponding residuals are given by

Humidity level 1: -2.05,-11.05, 4.45, 7.45, 1.2,

Humidity level 2: 19.15,-12.85,-12.35,-1.35, 7.4,

Humidity level 3: -1.45, 7.55,-0.95,-3.95,-1.2,

Humidity level 4: -15.65, 16.35, 8.85,-2.15,-7.4,

Humidity level 5: -2.05,-11.05, 4.45, 7.45, 1.2,

First order the residuals:

-15.65,-12.85,-12.35,-11.05,-7.4,-3.95,-2.15,-2.05,-1.45,-1.35,-1.2,-0.95

1.2, 4.45, 7.4, 7.45, 7.55, 8.85, 16.35, 19.15

Than find corresponding \(z\)scores:

-1.96,-1.44,-1.15,-0.93,-0.76,-0.60,-0.45,-0.32,-0.19,-0.06,

0.06,0.19,0.32,0.45,0.60,0.76,0.93,1.15,1.44,1.96.

This is given so you know how to compute residuals yourself, which is quite important the following normal probability plot suggest that the assumption of normality is plausible (by looking into the line it is fairly linear).

The assumption of equal variances can be checked by looking into plot of residuals versus fitted values. By the formula, the fitted values are:

\(\begin{aligned}{l}{\rm{For j = 1: 687}}{\rm{.05,733}}{\rm{.05,728}}{\rm{.55,803}}{\rm{.55,826}}{\rm{.8,}}\\{\rm{For j = 2: 772}}{\rm{.85,818}}{\rm{.85,814}}{\rm{.35,889}}{\rm{.35,912}}{\rm{.6,}}\\{\rm{For j = 3: 839}}{\rm{.45,885}}{\rm{.45,880}}{\rm{.95,955}}{\rm{.95,979}}{\rm{.2,}}\\{\rm{For j = 4: 890}}{\rm{.65,936}}{\rm{.65,932}}{\rm{.15,1007}}{\rm{.15,1030}}{\rm{.4}}{\rm{.}}\end{aligned}\)

Since there are a lot of fitted values (20), the computation is not shown. For example, for\(j = 2{\rm{\;and\;}}i = 5{\rm{,\;}}\), you have

\({\hat x_{52}} = {\bar x_{5.}} + {\bar x_{.2}} - {\bar x_{..}} = 841.6 + 843.5 - 866.25 = 818.85{\rm{,\;}}\)

Which is the only value marked blue.

From the plotted residuals versus fitted values you can conclude that the assumption of equal variances stand, because the plot does not show any pattern. The plot was created by plotting fitted value with corresponding residual!