Question

1. The critical *F* value with 6 numerator and 40
denominator degrees of freedom at *α* = .05 is

2. Consider the following information.

SSTR = 6750 | H_{0}: μ_{1}
=μ_{2} =μ_{3}
=μ_{4} = μ_{5} |

SSE = 8000 | H_{a}: At least one mean is different |

The null hypothesis is to be tested at the 5% level of
significance. The *p*-value is

Answer #1

**hypothesis**:-

at least one mean is different.

*given data
are:-*

**the test
statistic is :-**

**the p value =
0.0003**

[ in any blank cell of excel type =F.DIST.RT(5.625,6,40) ]

**decision**:-

p value = 0.0003 <0.05 (alpha)

so, we *reject the
null hypothesis .*

*** if you have any doubt regarding the problem please write it
in the comment box.if you are satisfied please give me a
**LIKE** if possible...

Consider the following ANOVA experiments. (Round your answers to
two decimal places.)
(a) Determine the critical region and critical value that are
used in the classical approach for testing the null hypothesis
H0: μ1 =
μ2 = μ3 =
μ4, with n = 19 and α =
0.05.
F ≥
(b) Determine the critical region and critical value that are used
in the classical approach for testing the null hypothesis
H0: μ1 =
μ2 = μ3 =
μ4 = μ5,...

The following data were obtained for a randomized block design
involving five treatments and three blocks: SST = 510, SSTR = 370,
SSBL = 95. Set up the ANOVA table. (Round your value for F
to two decimal places, and your p-value to three decimal
places.)
Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F
p-value
Treatments
Blocks
Error
Total
Test for any significant differences. Use α = 0.05.
State the null and alternative hypotheses.
H0: Not...

SSTR = 6,750
H0:
μ1=μ2=μ3=μ4
SSE = 8,000
Ha: at least one mean is different
nT = 20
Refer to Exhibit 10-11. The null hypothesis
Question 15 options:
should be rejected
should not be rejected
was designed incorrectly
None of these alternatives is correct.

SSTR = 6,750
SSE = 8,000
nT = 20
H0: μ1 = μ2 = μ3 =
μ4
Ha: Not all population means are equal
a. The mean square between treatments (MSTR) equals
b. The mean square within treatments (MSE) equals
c. The test statistic to test the null hypothesis equals
d. The critical F value at alpha = 0.05 is
e. The null hypothesis is

The China Health and Nutrition Survey aims to examine the
effects of the health, nutrition, and family planning policies and
programs implemented by national and local governments. It, for
example, collects information on number of hours Chinese parents
spend taking care of their children under age 6. The side-by-side
box plots below show the distribution of this variable by
educational attainment of the parent. Also provided below is the
ANOVA output for comparing average hours across educational
attainment categories.
Df...

1. An F test with five degrees of freedom in the
numerator and seven degrees of freedom in the denominator produced
a test statistic whose value was 3.97.
What is the P-value if the test is one-tailed? Round
the answer to three decimal places.
2. A hypothesis test is performed, and the P-value is
0.03. State whether the following statements are true or false.
H0 is rejected at the 5% level.
H0 is rejected at the 2% level.
H0 is...

You may need to use the appropriate technology to answer this
question.
An experiment has been conducted for four treatments with seven
blocks. Complete the following analysis of variance table. (Round
your values for mean squares and F to two decimal places,
and your p-value to three decimal places.)
Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F
p-value
Treatments
300
Blocks
700
Error
100
Total
1,100
Use α = 0.05 to test for any significant
differences....

An experiment has been conducted for four treatments with seven
blocks. Complete the following analysis of variance table. (Round
your values for mean squares and F to two decimal places,
and your p-value to three decimal places.)
Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F
p-value
Treatments
900
Blocks
200
Error
Total
1,600
Use α = 0.05 to test for any significant
differences.
State the null and alternative hypotheses.
H0: At least two of the population...

Use the following information to answer questions
9-15.
The length of time spent in the examination room is recorded for
each patient seen by each physician at an orthopedic clinic.
Does the data provide a significance difference in mean
times?
Physician 1
Physician 2
Physician 3
Physician 4
34
33
17
28
25
35
30
33
27
31
30
31
31
31
26
27
26
42
32
32
34
33
28
33
21
26
40
29
X.1=198
X. 2=205...

A study of the properties of metal plate-connected trusses used
for roof support yielded the following observations on axial
stiffness index (kips/in.) for plate lengths 4, 6, 8, 10, and 12
in:
4: 342.2 409.5 311.0 326.5 316.8 349.8 309.7
6: 420.1 347.2 361.0 404.5 331.0 348.9 381.7
8: 398.4 366.2 351.0 357.1 409.9 367.3 382.0 10: 366.7 452.9
461.4 433.1 410.6 384.2 362.6
12: 417.4 441.8 419.9 410.7 473.4 441.2 465.8
Does variation in plate length have any effect...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 20 minutes ago

asked 24 minutes ago

asked 24 minutes ago

asked 24 minutes ago

asked 25 minutes ago

asked 31 minutes ago

asked 34 minutes ago

asked 34 minutes ago

asked 42 minutes ago

asked 47 minutes ago

asked 50 minutes ago