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 |

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.

*H*_{0}: *μ*_{x} =
*μ*_{y} = *μ*_{z}

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

*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.

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 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 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 sufficient
evidence to conclude that the mean absorbency ratings for the three
brands are not all equal.

Answer #1

Excel:

Data --> Data analysis --> Anova: Single Factor

Output:

ANOVA | ||||||

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

Between Groups | 416 | 2 | 208 | 6.077922 | 0.021363 | 4.256495 |

Within Groups | 308 | 9 | 34.22222 | |||

Total | 724 | 11 |

**Answers:**

*H*_{0}: *μ*_{x} =
*μ*_{y} = *μ*_{z}

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

test statistic = 6.08

*p*-value = 0.021

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

(Reason : since *p*-value = 0.021 < 0.05, we reject
H0)

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....

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....

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...

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...

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 =...

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
assembled correctly with each method, 39 employees were randomly
selected and randomly assigned to the three proposed methods in
such a way that each method was used by 13 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 = 13,490; SSTR =...

In an experiment designed to test the output levels of three
different treatments, the following results were obtained: SST =
320, SSTR = 130,
nT = 19.
Set up the ANOVA table. (Round your values for MSE and
F to two decimal places, and your p-value to four
decimal places.)
Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F
p-value
Treatments
Error
Total
Test for any significant difference between the mean output
levels of the three treatments....

You may need to use the appropriate technology to answer this
question.
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
106
93
119
113
83
113
126
84
106
103
101
132
107
90
114
109
116
129
98
111
103
115
119
105
97
78
116
xj
119
106
101
sj2
149.14
155.33
187.56
State the null...

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