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27 . The predicted value for y is:
$\widehat{y}$ = 2.3 – 0.1(4.1) = 1.89. The value of 2.32 is more than two standard deviations from the predicted value, so it qualifies as an outlier.
Residual for (4.1, 2.34): 2.32 – 1.89 = 0.43 (0.43>2(0.13))
28 .
29 .
H
_{0} :
μ 1 =
μ 2 =
μ 3 =
μ 4
H
_{a} : At least two of the group means
μ 1,
μ 2,
μ 3,
μ 4 are not equal.
30 . The independent samples t -test can only compare means from two groups, while one-way ANOVA can compare means of more than two groups.
31 . Each sample appears to have been drawn from a normally distributed populations, the factor is a categorical variable (method), the outcome is a numerical variable (test score), and you were told the samples were independent and randomly selected, so those requirements are met. However, each sample has a different standard deviation, and this suggests that the populations from which they were drawn also have different standard deviations, which is a violation of an assumption for one-way ANOVA. Further statistical testing will be necessary to test the assumption of equal variance before proceeding with the analysis.
32 . One of the assumptions for a one-way ANOVA is that the samples are drawn from normally distributed populations. Since two of your samples have an approximately uniform distribution, this casts doubt on whether this assumption has been met. Further statistical testing will be necessary to determine if you can proceed with the analysis.
33 . SS _{within} is the sum of squares within groups, representing the variation in outcome that cannot be attributed to the different feed supplements, but due to individual or chance factors among the calves in each group.
34 . SS _{between} is the sum of squares between groups, representing the variation in outcome that can be attributed to the different feed supplements.
35 .
k = the number of groups = 4
n
_{1} = the number of cases in group 1 = 30
n = the total number of cases = 4(30) = 120
36 .
SS
_{total} =
SS
_{within} +
SS
_{between} so
SS
_{between} =
SS
_{total} –
SS
_{within}
621.4 – 374.5 = 246.9
37 . The mean squares in an ANOVA are found by dividing each sum of squares by its respective degrees of freedom (
df ).
For
SS
_{total} ,
df =
n – 1 = 120 – 1 = 119.
For
SS
_{between} ,
df = k – 1 = 4 – 1 = 3.
For
SS
_{within} ,
df = 120 – 4 = 116.
MS
_{between} =
$\frac{246.9}{3}$ = 82.3
MS
_{within} =
$\frac{374.5}{116}$ = 3.23
38 . $F=\frac{M{S}_{between}}{M{S}_{within}}=\frac{82.3}{3.23}=25.48$
39 . It would be larger, because you would be dividing by a smaller number. The value of MS _{between} would not change with a change of sample size, but the value of MS _{within} would be smaller, because you would be dividing by a larger number ( df _{within} would be 136, not 116). Dividing a constant by a smaller number produces a larger result.
40 . All but choice c, –3.61. F Statistics are always greater than or equal to 0.
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