Ttest for the slope:
t
= 26
.
02
/
3
.
15 = 8
.
26 on df= 13. We
get a pvalue
<
2
*
.
001. Strong evidence that the slope is
>
0.
(d) prediction of one new value: 52
.
37+26
.
02
*
19 = 546
.
87
thousands of eggs if the female weighs 19kg, and 52
.
37 +
26
.
02
*
24 = 677
.
0 thousands of eggs at 24kg.
Larger. This is because ¯
x
= 16
.
1kg so 19kg is closer to the
mean female’s weight than 24kg. The SE of the prediction
is larger when (
x

¯
x
)
2
is larger, and here (24

¯
x
)
2
is larger
than (19

¯
x
)
2
.
3. See Ley
et al.
Nature 444:1022 (2006).
(a) Paired ttest, 2 sided.
(b)
t
= 3
.
09, df= 12

1 = 11 so the pvalue is 2
*
.
005
< p <
2
*
.
01 i.e.
.
01
< p < .
02. We reject the null hypothesis. In
obese subjects (such as those sampled for the study), the
mean bacteroidetes abundance is higher at week 54 than at
week 12 of the diet.
4.(a) We can get SSError as (8

1)
*
785
2
+
· · ·
+(8

1)
*
62
2
=
4
,
427
,
213.
To get SSCultivar, we can first get the grand
mean (8
*
1830+
· · ·
+8
*
403)
/
48 = 635
.
17 then SSCultivar is
8
*
(1830

635
.
17)
2
+
· · ·
+8
*
(403

635
.
17)
2
= 13
,
962
,
803.
We can also get SSCultivar by subtracting SSError from
SSTotal.
Source
df
SS
MS
F
pvalue
Cultivar
5
13,962,803
2,792,561
26.5
< .
001
Error
42
4,427,213
105,409.8
Total
47
18,390,020
No, this analysis does not provide evidence that the 5 oca
cultivars do not all have the same mean. The significant p
value could be due to the rhubarb having a different mean
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 Fall '08
 Staff
 Normal Distribution, OCA, Normal scores plot, oca cultivars, female’s weight

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