Question

Pr d) OBTAIN A 9 CONFIDENCE INTERVAL FOR THİS CONTRAST.

Days Status
29.0 1
42.0 1
38.0 1
40.0 1
43.0 1
40.0 1
30.0 1
42.0 1
30.0 2
35.0 2
39.0 2
28.0 2
31.0 2
31.0 2
29.0 2
35.0 2
29.0 2
33.0 2
26.0 3
32.0 3
21.0 3
20.0 3
23.0 3
22.0 3

Answer 1

(a)

From above plot it is observed that normality assumption is valid.

From the above tests it is observed that the equality of variances is valid.

$(b)~\hat{\mu}_i=\bar{y}_i=\frac{1}{n_i}\sum_{i=1}^{n_i}y_{ij}.\\ Here~n_1=8,~ n_2=10, ~n_3=6.\\ \hat{\mu}_1= 38.000,~\hat{\mu}_2=32.000,~\hat{\mu}_3=24.000\\ Standard~error~of~\hat{\mu}_1= 5.477\\ Standard~error~of~\hat{\mu}_2= 3.464\\ Standard~error~of~\hat{\mu}_3=4.427\\$

(c)

One-way ANOVA: Days versus Status

Source DF SS MS F P
Status 2 672.0 336.0 16.96 0.000
Error 21 416.0 19.8
Total 23 1088.0

Since p-value<0.05 so we conclude that at least one mean is different from others.

(d)

$90\%~Confidence~interval~for~\mu_1-\mu_3~is\\ \left(\hat{\mu}_1-\hat{\mu}_3-t_{0.05,21}\times \sqrt{MSE\left(\frac{1}{8}&plus;\frac{1}{6} \right )},~ \hat{\mu}_1-\hat{\mu}_3&plus;t_{0.05,21}\times \sqrt{MSE\left(\frac{1}{8}&plus;\frac{1}{6} \right )}\right )\\ =(9.8649,~18.1351)\\ where,~t_{0.05,21}=1.7207\\ Since~the~confidence~interval~does~not~contain~zero~hence\\~\mu_1~is~different~from~\mu_3.\\ (e)~\hat{L}=\frac{\hat{\mu}_1&plus;3\hat{\mu}_3}{4}-\hat{\mu}_2=-4.5\\\\ Var(\hat{L})=\left(\frac{1}{16n_1}&plus;\frac{9}{16n_3}&plus;\frac{1}{n_2} \right )\sigma^2\\ \hat{se}(\hat{L})=estimate~of~standard~error~of~\hat{L}\\ =\sqrt{(1/(16*8)&plus;9/(16*6)&plus;1/10)*MSE}\\ =\sqrt{(1/(16*8)&plus;9/(16*6)&plus;1/10)* 19.8}=1.9977\\ t_{0.005,21}=2.8314\\ 99\%~C.I.~for~L~is\\ \left(\hat{L}-t_{0.005,21}\hat{se}(\hat{L}),~ \hat{L}&plus;t_{0.005,21}\hat{se}(\hat{L})\right )=(-10.1563,~1.15629)$