Question

In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equal-size towels were used, with four sections of towels tested per brand. The absorbency rating data follow.

Brand | ||
---|---|---|

x |
y |
z |

91 | 100 | 84 |

99 | 95 | 88 |

89 | 93 | 90 |

85 | 104 | 74 |

At a 0.05 level of significance, does there appear to be a difference in the ability of the brands to absorb water?

State the null and alternative hypotheses.

*H*_{0}: Not all the population means are
equal.

*H*_{a}: *μ*_{x} =
*μ*_{y} =
*μ*_{z}*H*_{0}:
*μ*_{x} = *μ*_{y} =
*μ*_{z}

*H*_{a}: *μ*_{x} ≠
*μ*_{y} ≠
*μ*_{z}
*H*_{0}: At least two of the population means are
equal.

*H*_{a}: At least two of the population means are
different. *H*_{0}: *μ*_{x} =
*μ*_{y} = *μ*_{z}

*H*_{a}: Not all the population means are equal.
*H*_{0}: *μ*_{x} ≠
*μ*_{y} ≠ *μ*_{z}

*H*_{a}: *μ*_{x} =
*μ*_{y} = *μ*_{z}

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the *p*-value. (Round your answer to three decimal
places.)

*p*-value =

State your conclusion.

Reject *H*_{0}. There is not sufficient evidence
to conclude that the mean absorbency ratings for the three brands
are not all equal. Do not reject *H*_{0}. There is
not sufficient evidence to conclude that the mean absorbency
ratings for the three brands are not all
equal. Reject *H*_{0}. There
is sufficient evidence to conclude that the mean absorbency ratings
for the three brands are not all equal. Do not reject
*H*_{0}. There is sufficient evidence to conclude
that the mean absorbency ratings for the three brands are not all
equal.

Answer #1

For the given data set anova testing can be done in excel tool.

The hypotheses are:

H_{0}: μ_{x} = μ_{y} =
μ_{z}

H_{a}: Not all the population means are equal.

Anova: Single Factor | ||||||

SUMMARY | ||||||

Groups |
Count |
Sum |
Average |
Variance |
||

X | 4 | 364 | 91 | 34.66667 | ||

Y | 4 | 392 | 98 | 24.66667 | ||

Z | 4 | 336 | 84 | 50.66667 | ||

ANOVA | ||||||

Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |

Between Groups | 392 | 2 | 196 | 5.345455 | 0.029505 | 4.256495 |

Within Groups | 330 | 9 | 36.66667 | |||

Total | 722 | 11 |

The test statistic:

F=5.35

P-value:

P-value calculated for F statistic as:

P-value=0.0295

Conclusion:

Reject *H*_{0}. There is sufficient
evidence to conclude that the mean absorbency ratings for the three
brands are not all equal.

Since P-value< 0.05.

In a completely randomized experimental design, three brands of
paper towels were tested for their ability to absorb water.
Equal-size towels were used, with four sections of towels tested
per brand. The absorbency rating data follow.
Brand
x
y
z
90
99
83
100
97
87
87
93
90
95
99
72
At a 0.05 level of significance, does there appear to be a
difference in the ability of the brands to absorb water?
State the null and alternative hypotheses....

In a completely randomized experimental design, three brands of
paper towels were tested for their ability to absorb water.
Equal-size towels were used, with four sections of towels tested
per brand. The absorbency rating data follow.
Brand
x
y
z
92
98
84
101
95
87
88
93
88
87
102
77
At a 0.05 level of significance, does there appear to be a
difference in the ability of the brands to absorb water?
State the null and alternative hypotheses....

The following data were obtained for a randomized block design
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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...

Consider the experimental results for the following randomized
block design. Make the calculations necessary to set up the
analysis of variance table.
Treatments
A
B
C
Blocks
1
10
9
8
2
12
6
5
3
18
16
14
4
20
18
18
5
8
7
8
Use α = 0.05 to test for any significant
differences.
State the null and alternative hypotheses.
H0: Not all the population means are
equal.
Ha: μA =
μB = μC
H0: μA =...

Develop the analysis of variance computations for the following
completely randomized design. At α = 0.05, is there a
significant difference between the treatment means?
Treatment
A
B
C
136
108
91
119
115
81
113
125
85
106
105
102
130
108
88
115
109
117
129
96
109
112
115
121
104
99
85
107
xj
120
107
100
sj2
110.29
119.56
186.22
State the null and alternative hypotheses.
H0: At least two of the population means are...

You may need to use the appropriate technology to answer this
question.
Consider the experimental results for the following randomized
block design. Make the calculations necessary to set up the
analysis of variance table.
Treatments
A
B
C
Blocks
1
10
9
8
2
12
7
5
3
18
15
14
4
20
18
18
5
8
7
8
Use α = 0.05 to test for any significant
differences.
State the null and alternative hypotheses.
H0: μA ≠ μB ≠...

Three different methods for assembling a product were proposed
by an industrial engineer. To investigate the number of units
assembled correctly with each method, 42 employees were randomly
selected and randomly assigned to the three proposed methods in
such a way that each method was used by 14 workers. The number of
units assembled correctly was recorded, and the analysis of
variance procedure was applied to the resulting data set. The
following results were obtained: SST = 13,960; SSTR =...

Three different methods for assembling a product were proposed
by an industrial engineer. To investigate the number of units
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In an experiment designed to test the output levels of three
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nT = 19.
Set up the ANOVA table. (Round your values for MSE and
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Source
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Sum
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Degrees
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Mean
Square
F
p-value
Treatments
Error
Total
Test for any significant difference between the mean output
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To study the effect of temperature on yield in a chemical
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Temperature
50°C
60°C
70°C
33
30
22
24
30
28
35
35
27
38
24
30
30
21
28
Construct an analysis of variance table. (Round your values for
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Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F...

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