Question
In: Statistics and Probability
Block Treatment 1 2 3 4 Treatment Mean Tr1 2 1 2 3 2 Tr2 4...
| Block | |||||
| Treatment | 1 | 2 | 3 | 4 | Treatment Mean |
| Tr1 | 2 | 1 | 2 | 3 | 2 |
| Tr2 | 4 | 4 | 1 | 1 | 2.5 |
| Tr3 | 3 | 4 | 3 | 2 | 3 |
| Block Mean | 2 | 3 | 3 | 2 | overall mean = 2.5 |
Consider the randomized block design with 4 blocks and 3 treatments given above. Test H0: there is no difference between treatment effects at α = .05.
Solutions
orchestra answered 1 year agoRelated Solutions
Block Treatment 1 2 3 4 Treatment Mean Tr1 2 1 2 3 2 Tr2 4...
Block
Treatment
1
2
3
4
Treatment Mean
Tr1
2
1
2
3
2
Tr2
4
4
1
1
2.5
Tr3
3
4
3
2
3
Block Mean
2
3
3
2
overall mean = 2.5
Consider the randomized block design with 4 blocks and 3
treatments given above. What is the value of the F
statistic for blocks?
Treatment 1 Treatment 2 Treatment 3 0
Treatment 1
Treatment 2 Treatment 3
0
2
```4
0
3
2 G = 36
0
1
4
ΣX2 = 114
3
3
3
0
2
4
0
1
4
––––––––––––––––––––––––––––––––––––
T =
3
T =
12
T = 21
SS =
7.5
SS =
4
SS = 3.5
1. Use an analysis of variance with α = .05 to determine whether
these data indicate any significant mean differences among the
treatments.
2. Compute η2,...
SE-FamilySize 1 1 4 3 2 4 2 3 4 2 4 1 4 2 2...
SE-FamilySize
1
1
4
3
2
4
2
3
4
2
4
1
4
2
2
4
5
4
5
4
4
2
4
3
1
2
3
5
5
5
Make a confidence interval. Be sure you show all the steps you
took. Include a screen shot of any applet you used in your
calculations.
2. Choose a confidence level (1 – α).
3. What is xbar?
4. What is s?
5. What is t? (Show a screen shot...
1. sign and symptoms 2. investigation 3. treatment 4. complication 5. last study
(I want lika a persentaiton ) About (Topic: peptic ulcer)
1. sign and symptoms
2. investigation
3. treatment
4. complication
5. last study
Question 4: Consider the following sample information for three treatment groups. Treatment 1 Treatment 2 Treatment...
Question 4:
Consider the following sample information for three treatment
groups.
Treatment 1 Treatment 2 Treatment 3
$8 $3 $3
6 2 4
10 4 5
9 3 4
Using the 0.05 significance level, use the ANOVA method to test
the null hypothesis that the three treatment means are equal.
Consider the following sample information for three treatment groups: Treatment 1 Treatment 2 Treatment 3 15...
Consider the following sample information for three treatment
groups:
Treatment 1
Treatment 2
Treatment 3
15
18
11
18
21
12
13
17
15
12
15
9
(a) using the 0.05 significance level, test the null hypothesis
that the three treatment means are equal
(b) Interpret your results.
PLEASE SHOW WORK! IM TRYING TO LEARN!!!!
Consider the following sample information for three treatment groups: Treatment 1 Treatment 2 Treatment 3 15...
Consider the following sample information for three treatment
groups:
Treatment 1
Treatment 2
Treatment 3
15
18
11
18
21
12
13
17
15
12
15
9
(a) Using the 0.05 significance level, test the null hypothesis
that the three treatment means are equal.
(b) Interpret your results.
A 2^3 design is confounded in 4 blocks of size 2. The principal block contains treatments...
A 2^3 design is confounded in 4 blocks of size 2. The principal
block contains treatments (1) and abc. Another block contains a.
What is the second element of that block. (b, c, ab, ac, or
bc?)
MONTH ZEMIN CORP. MARKET 1 7% 4% 2 2% 1% 3 2% 2% 4 -3% -1%...
MONTH ZEMIN CORP. MARKET
1 7% 4%
2 2% 1%
3 2% 2%
4 -3% -1%
5 4% 2%
6 2% 3%
a. Given the holding-period returns shown above, compute the
average returns and the standard deviations for the Zemin
Corporation and for the market.
b. If Zemin's beta is 1.04 and the risk-free rate is 9
percent, what would be an appropriate required return for an
investor owning Zemin? (Note: Because the returns of Zemin
Corporation are based on...
Question 5 a) (1) X~Normal(mean=4, standard deviation=3), (2) Y~Normal(mean=6, standard deviation = 4), and (3) X...
Question 5
a) (1)
X~Normal(mean=4, standard deviation=3), (2) Y~Normal(mean=6,
standard deviation = 4), and (3) X and Y are independent, then,
P(X+Y>13) equals (in 4 decimal places)
Answers options: a) 0.7257, b) 0.3341, c) 0.2743, d) 0.6759, e)
none of these
b) Let X~Gamma(4, 1.2). Which of the following
is possible R code for computing the probability that X <
2.6?
Answers options: a) dgam(2.6, 4, 1.2), b) pgamma(4, 1.2, 2.6),
c) dgamma(2.6, 4, 1.2), d) pgamma(2.6, 4, 1.2), e)...
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