EBK THE BASIC PRACTICE OF STATISTICS
EBK THE BASIC PRACTICE OF STATISTICS
7th Edition
ISBN: 8220103935319
Author: Moore
Publisher: YUZU
Question
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Chapter 30, Problem 30.29E

a.

To determine

The contrast expression for the comparison of tropical flowers.

a.

Expert Solution
Check Mark

Answer to Problem 30.29E

The First hypothesis becomes:

H0:L1=0vsHa:L10

The second hypothesis becomes:

H0:L2=0vsHa:L20

The third Hypothesis becomes:

H0:L3=0vsHa:L30

Explanation of Solution

Given info:

The data represents the comparisons of the tropical flowers. The variables are the lenghts aof bihai variety of Heliconia that is represented in two forms based on the color of caribaea variety.

Calculation:

In the ANOVA construction, the comparison of means μred,μbihai,andμyellow of the population, a population contrast is a combination of means.

L=credμred+cbihaiμbihai+cyellowμyellow

Group μbihai indicates the bihai variety of Heliconia, group μred indicates the red variety of Caribaea and group μyellow denotes the yellow variety of Caribaea.

The three hypotheses are terms of three contrasts are as follows:

The first hypothesis in terms of three contrasts is,

L1=cbihaiμbihaicredμred=(1)μbihai+(1)μred+(0)μyellow

Thus, The First hypothesis becomes:

H0:L1=0vsHa:L10

The second hypothesis in terms of three contrasts is,

L2=cbihaiμbihaicyellowμyellow=(0)μred+(1)μbihai+(1)μyellow

Thus, The second hypothesis becomes:

H0:L2=0vsHa:L20

The third hypothesis in terms of three contrasts is,

L3=cbihaiμbihai+credμred+cyellowμyellow=(1)μbihai(μred2+μyellow2)=(1)μbihai+(12)μred+(12)μyellow

Thus, The third Hypothesis becomes:

H0:L3=0vsHa:L30

b.

To determine

The null and alternative hypothesis and the sample contrast and also assess its statistical significance.

b.

Expert Solution
Check Mark

Answer to Problem 30.29E

The First hypothesis becomes,

H0:L1=0vsHa:L10

The second hypothesis becomes,

H0:L2=0vsHa:L20

The third Hypothesis becomes,

H0:L3=0vsHa:L30

The first sample contrast is,

L^1=7.887,SE1=0.466

The second sample contrast is,

L^2=11.418,SE2=0.517

The third sample contrast is,

L^3=29.495,SE3=0.4319

The interpretation is that , there is a evidence that the mean bihai variety of Heliconia differs from the average of the two forms of Caribaea.

Explanation of Solution

Calculation:

In the ANOVA construction, the comparison of means μred,μbihai,andμyellow of the population, a population contrast is a combination of means.

L=credμred+cbihaiμbihai+cyellowμyellow (1)

The sample contrast L^ has a standard error is calculated as:

SEL^=spcred2n1+cbihai2n2+cyellow2n3

Software procedure:

Step by step procedure to obtain two way ANOVA using the MINITAB software:

  • Choose Stat > ANOVA > One-way.
  • In Response, select the column of Length.
  • In Factor, select the column of Variety
  • Click on Comparison> Tick on Tukey under Comparison procedures assuming equal variances and then click OK.
  • Click OK.

The Minitab output is shown below.

EBK THE BASIC PRACTICE OF STATISTICS, Chapter 30, Problem 30.29E

Fig (1)

From the Minitab result, the pooled standard deviation sp=1.44 and the degree of freedom is 51.

The standard errors for the first sample contrast are,

SEL^1=spcred2n1+cbihai2n2+cyellow2n3=1.44(1)223+116+015=1.440.1046=0.466

Thus, The standard error for first sample contrast is 0.466.

Substitute 1 for cred , 39.711 for μred , 1 for cbihai , 47.598 for μbihai , 0 for cyellow , and 36.180 in the above equation (1).

L^1=credμred+cbihaiμbihai+cyellowμyellow=(1)×39.711+(1)×47.598+(0)×36.180=7.887

The standard error for second sample contrast is,

SEL^2=spcred2n1+cbihai2n2+cyellow2n3=1.44(0)223+116+(1)215=1.440.1291=0.517

Thus, The standard error for the second sample contrast is 0.5617.

Substitute 0 for cred , 39.711 for μred , 1 for cbihai , 47.598 for μbihai , (1) for cyellow , and 36.180 for μyellow in the above equation (1).

L^2=credμred+cbihaiμbihai+cyellowμyellow=(0)×39.711+(1)×47.598+(1)×36.180=11.418

The standard error for third sample contrast is,

SEL^3=spcred2n1+cbihai2n2+cyellow2n3=1.44(0.5)223+(1)216+(0.5)215=1.440.2523+116+0.2515=1.440.0899

Further, solve the above expression.

SEL^3=1.44×0.2999=0.4319

Thus, the standard error for third sample contrast is 0.4319.

Substitute (0.5) for cred , 39.711 for μred , 1 for cbihai , 47.598 for μbihai , (0.5) for cyellow , and 36.180 for μyellow in the above equation (1).

L^1=credμred+cbihaiμbihai+cyellowμyellow=(0.5)×39.711+(1)×47.598+(0.5)×36.180=0.01259+47.59818.09=29.495

The hypotheses are given as:

The First hypothesis becomes:

H0:L1=0vsHa:L10

The second hypothesis becomes:

H0:L2=0vsHa:L20

The third Hypothesis becomes:

H0:L3=0vsHa:L30

The t- statistics is,

t=L3^SE3=29.4950.4319=68.2912

The tabulated value with 51 degree of freedom is 1.6753 which is very much lesser than 68.2912. The null hypothesis is rejected for population contrast L3 . There is very strong evidence that the population contrast L3 is not equal to zero for the t-statistics with 51 degree of freedom. Therefore, μbihai and (μred+μyellow)2 significantly differ.

There is evidence that the mean bihai variety of Heliconia differs from the average of the two forms of Caribaea.

c.

To determine

The 90% confidence interval for the population contrast.

c.

Expert Solution
Check Mark

Answer to Problem 30.29E

The 90% confidence interval for L3 population contrast is (28.7716,30.218) .

Explanation of Solution

Calculation:

The 90% confidence interval from the population contrast for the population contrast is calculated as:

L3=μbihai(μred+μyellow2)

The 90% confidence interval for L3=29.495 uses, t=1.675(fromt-tablewith51df) .

L3^±tSE3=29.495±1.675(0.4319)

The lower limit is,

Lowerlimit=29.4951.675(0.4319)=28.7716

The upper limit is,

UpperLimit=29.495+1.675(0.4319)=30.218

The 90% confidence interval is (28.7716,30.218) for L3 population contrast.

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