Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 9.5, Problem 21E
To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer
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Answer to Problem 21E

Solution:

The confidence interval is (0.1824, 6.9828).

Explanation of Solution

Given Information:

From the given information the total number of flour bags filled by Machine A is n1=11 and by Machine B is n2=10. Their population variances are s12=0.584 and s22=0.499 at 99% level of confidence.

Concept:

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, σ12σ22.

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that σ12 is larger than σ22.

3. If both endpoints of the interval are less than 1 it concludes that σ12 is smaller than σ22

Formula used:

The formula to calculate the confidence interval is,

(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1α2))

The formula to calculate the point estimate is,

s12s22

Where, s1 and s2 are the sample variances for the respective samples.

For the given level of confidence c, the level of significance is α=1c.

The right and left endpoints for F-distribution are α2 and 1α2 respectively. The critical values are Fα2 and F1α2 that can be determined in the F-distribution table using the column for df1=n11 and row for df2=n21.

Calculation:

Substitute 0.584 for s12 and 0.499 for s22 in the point estimate formula as,

s12s22=0.5840.499=1.1703

At 99% level of confidence or c=0.99, the level of significance is,

α=10.99=0.01

The right endpoint for F-distribution is α2. Substitute 0.01 for α in α2.

α2=0.012=0.005

Similarly the left endpoint for F-distribution is 1α2. Substitute 0.01 for α in α2.

1α2=10.005=0.995

The degrees of freedom are,

df1=n11……(1)

And,

df2=n21……(2)

The sample sizes are given as n1=11 and n2=10.

Substitute 11 for n1 in equation (1).

df1=n11=111=10

Substitute 10 for n2 in equation (2).

df2=n21=101=9

For the column of df1=10 and the row of df2=9 in the F-distribution table under the area of α2, the value of Fα2 is 6.4172.

For the column of df1=10 and the row of df2=9 in the F-distribution table under the area of 1α2, the value of F1α2 is 0.1676.

The formula to calculate the confidence interval is,

(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1α2)).

Substitute 1.1703 for s12s22, 6.4172 for Fα/2 and 0.1676 for F(1α/2) in the above formula of confidence interval,

Confidence interval=(1.1703×16.4172)<σ12σ22<(1.1703×10.1676)=(0.1824)<σ12σ22<(6.9828)

Thus, the confidence interval is (0.1824, 6.9828).

Interpretation:

Since the value 1 is in the interval (0.1824, 6.9828), the data do not provide evidence at 99% level of confidence that, the population variances of the amount of flour per bag for Machine A and Machine B are unequal.

If it is important that the two machines have the same variances, the inspector needs to adjust Machine A as it has higher variance than Machine B, so the variance will be 0.499 for both the machines.

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Chapter 9 Solutions

Beginning Statistics, 2nd Edition

Ch. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.2 - Prob. 1ECh. 9.2 - Prob. 2ECh. 9.2 - Prob. 3ECh. 9.2 - Prob. 4ECh. 9.2 - Prob. 5ECh. 9.2 - Prob. 6ECh. 9.2 - Prob. 7ECh. 9.2 - Prob. 8ECh. 9.2 - Prob. 9ECh. 9.2 - Prob. 10ECh. 9.2 - Prob. 11ECh. 9.2 - Prob. 12ECh. 9.2 - Prob. 13ECh. 9.2 - Prob. 14ECh. 9.2 - Prob. 15ECh. 9.2 - Prob. 16ECh. 9.2 - Prob. 17ECh. 9.2 - Prob. 18ECh. 9.2 - Prob. 19ECh. 9.2 - Prob. 20ECh. 9.2 - Prob. 21ECh. 9.2 - Prob. 22ECh. 9.2 - Prob. 23ECh. 9.2 - Prob. 24ECh. 9.3 - Prob. 1ECh. 9.3 - Prob. 2ECh. 9.3 - Prob. 3ECh. 9.3 - Prob. 4ECh. 9.3 - Prob. 5ECh. 9.3 - Prob. 6ECh. 9.3 - Prob. 7ECh. 9.3 - Prob. 8ECh. 9.3 - Prob. 9ECh. 9.3 - Prob. 10ECh. 9.3 - Prob. 11ECh. 9.3 - Prob. 12ECh. 9.3 - Prob. 13ECh. 9.3 - Prob. 14ECh. 9.3 - Prob. 15ECh. 9.3 - Prob. 16ECh. 9.3 - Prob. 17ECh. 9.3 - Prob. 18ECh. 9.3 - Prob. 19ECh. 9.3 - Prob. 20ECh. 9.3 - Prob. 21ECh. 9.4 - Prob. 1ECh. 9.4 - Prob. 2ECh. 9.4 - Prob. 3ECh. 9.4 - Prob. 4ECh. 9.4 - Prob. 5ECh. 9.4 - Prob. 6ECh. 9.4 - Prob. 7ECh. 9.4 - Prob. 8ECh. 9.4 - Prob. 9ECh. 9.4 - Prob. 10ECh. 9.4 - Prob. 11ECh. 9.4 - Prob. 12ECh. 9.4 - Prob. 13ECh. 9.4 - Prob. 14ECh. 9.4 - Prob. 15ECh. 9.4 - Prob. 16ECh. 9.4 - Prob. 17ECh. 9.4 - Prob. 18ECh. 9.4 - Prob. 19ECh. 9.4 - Prob. 20ECh. 9.4 - Prob. 21ECh. 9.4 - Prob. 22ECh. 9.4 - Prob. 23ECh. 9.4 - Prob. 24ECh. 9.4 - Prob. 25ECh. 9.5 - Prob. 1ECh. 9.5 - Prob. 2ECh. 9.5 - Prob. 3ECh. 9.5 - Prob. 4ECh. 9.5 - Prob. 5ECh. 9.5 - Prob. 6ECh. 9.5 - Prob. 7ECh. 9.5 - Prob. 8ECh. 9.5 - Prob. 9ECh. 9.5 - Prob. 10ECh. 9.5 - Prob. 11ECh. 9.5 - Prob. 12ECh. 9.5 - Prob. 13ECh. 9.5 - Prob. 14ECh. 9.5 - Prob. 15ECh. 9.5 - Prob. 16ECh. 9.5 - Prob. 17ECh. 9.5 - Prob. 18ECh. 9.5 - Prob. 19ECh. 9.5 - Prob. 20ECh. 9.5 - Prob. 21ECh. 9.5 - Prob. 22ECh. 9.5 - Prob. 23ECh. 9.5 - Prob. 24ECh. 9.5 - Prob. 25ECh. 9.5 - Prob. 26ECh. 9.5 - Prob. 27ECh. 9.5 - Prob. 28ECh. 9.CR - Prob. 1CRCh. 9.CR - Prob. 2CRCh. 9.CR - Prob. 3CRCh. 9.CR - Prob. 4CRCh. 9.CR - Prob. 5CRCh. 9.CR - Prob. 6CRCh. 9.CR - Prob. 7CRCh. 9.CR - Prob. 8CRCh. 9.CR - Prob. 9CRCh. 9.CR - Prob. 10CRCh. 9.CR - Prob. 11CRCh. 9.CR - Prob. 12CRCh. 9.CR - Prob. 13CRCh. 9.CR - Prob. 14CRCh. 9.CR - Prob. 15CRCh. 9.P - Prob. 1P
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