Consider the following information.
SSTR = 6900
H0: μ1
=μ2 =μ3
=μ4
SSE = 8800
Ha: At least one mean is different
The mean square due to treatments (MSTR) equals...
a. 2300.
b. 400.
c.1687.5.
d. 2250.
If n = 5 (same for all treatments), the mean square due
to error (MSE) equals...
a. 400.
b. 2250.
c. 500.
d. 550
The test statistic to test the null hypothesis equals...
a. .22.
b. 4.50.
c. 4.22.
d. 4.18...
Consider the following information.
SSTR = 6900
H0: μ1
=μ2 =μ3
=μ4
SSE = 8800
Ha: At least one mean is different
The mean square due to treatments (MSTR) equals...
a. 2300.
b. 400.
c.1687.5.
d. 2250.
If n = 5 (same for all treatments), the mean square due
to error (MSE) equals...
a. 400.
b. 2250.
c. 500.
d. 550
The test statistic to test the null hypothesis equals...
a. .22.
b. 4.50.
c. 4.22.
d. 4.18...
Consider the following hypothesis test. H0: μ1 − μ2 = 0 Ha: μ1 −
μ2 ≠ 0 The following results are from independent samples taken
from two populations assuming the variances are unequal. Sample 1
Sample 2 n1 = 35 n2 = 40 x1 = 13.6 x2 = 10.1 s1 = 5.8 s2 = 8.2
(a) What is the value of the test statistic? 2.153 correct
(b) What is the degrees of freedom for the t distribution?
(Round your answer...
Consider the following hypothesis test.
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
The following results are from independent samples taken from
two populations.
Sample 1
Sample 2
n1 = 35
n2 = 40
x1 = 13.6
x2 = 10.1
s1 = 5.9
s2 = 8.1
(a)
What is the value of the test statistic? (Use
x1 − x2.
Round your answer to three decimal places.)
(b)
What is the degrees of freedom for the t...
Find the critical value from the Studentized range distribution
for H0: μ1 = μ2 = μ3 = μ4 = μ5, with n = 35 at α = 0.01. Round to
the nearest 3 decimal places.
A researcher studying four different
populations proposes the following:
H0 :
μ1=μ2=μ3=μ4
H1 : not all μi
are equal (i=1,2,3,4 )
He takes samples from each of the populations, obtaining the
following data:
Group 1
Group 2
Group 3
Group 4
Sample mean
110
105
91
102
Sample size
8
9
5
6
The within-group variation of the data (SSW) is equal to 108.
Using the Tukey-Kramer procedure, calculate the critical range
for making...
The null and alternate hypotheses are:
H0 : μ1 =
μ2
H1 : μ1 ≠
μ2
A random sample of 11 observations from one population revealed
a sample mean of 23 and a sample standard deviation of 4.6. A
random sample of 8 observations from another population revealed a
sample mean of 28 and a sample standard deviation of 3.6.
At the 0.05 significance level, is there a difference between
the population means?
State the decision rule. (Negative values should...
The null and alternate hypotheses are:
H0 : μ1 =
μ2
H1 : μ1 ≠
μ2
A random sample of 12 observations from one population revealed
a sample mean of 25 and a sample standard deviation of 3.5. A
random sample of 9 observations from another population revealed a
sample mean of 30 and a sample standard deviation of 3.5.
At the 0.01 significance level, is there a difference between
the population means?
State the decision rule. (Negative values should...
The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2
A random sample of 9 observations from one population revealed a
sample mean of 22 and a sample standard deviation of 3.9. A random
sample of 9 observations from another population revealed a sample
mean of 27 and a sample standard deviation of 4.1. At the 0.01
significance level, is there a difference between the population
means? State the decision rule. (Negative amounts should...
The null and alternate hypotheses are:
H0 : μ1 =
μ2
H1 : μ1 ≠
μ2
A random sample of 11 observations from one population revealed
a sample mean of 23 and a sample standard deviation of 1.1. A
random sample of 4 observations from another population revealed a
sample mean of 24 and a sample standard deviation of 1.3.
At the 0.05 significance level, is there a difference between
the population means?
State the decision rule. (Negative amounts should...