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 8.CR, Problem 9CR
To determine

To interpret:

A 95% confidence interval for the population variance for the volumes of soda in all the soda cans that come off that particular assumbly line.

Expert Solution & Answer
Check Mark

Answer to Problem 9CR

Solution:

The confidence interval for the population variance is given by (0.001, 0.003)

Explanation of Solution

Steps need to be followed while calculating the confidence interval.

STEP 1:

Find the point estimate, s2 for confidence interval for the population variance and s for confidence interval for the population standard deviation.

STEP 2:

Calculate α2 and 1-α2, based on the level of confidence c given.

STEP 3:

Find the critical value χα22 and χ(1-α2)2 for the distribution with n-1 degrees of freedom using the χ2-distribution table.

STEP 4:

Find the confidence interval for the population variance by substituting the necessary values in the formula

n-1s2χα22<σ2<n-1s2χ1-α22

Find the confidence interval for the population standard deviation by substituting the necessary values in the formula

n-1s2χα22<σ<n-1s2χ1-α22

Calculation:

The confidence interval for the population variance is given by (0.001, 0.003).

The volumes of soda in all the soda cans that come off that particular assumbly line will have the variance between 0.001 and 0.003.

Level confidence =95%

First let us calculate the sample variance for the given data.

The sample variance of a data having ‘n’ number of data values in the sample with mean ‘x-’ is given by

s2=xi-x-2n-1

Here, we need to find the mean ‘x-

x-=xin

=12.01+12.02+11.95+11.99+11.94+12.01+12.03+11.98+12.00+12.03+11.98+12.05+11.93+11.9814

x-=167.914

x-=11.9929

x-=12

Now, construct a table of deviations and squared deviations of the data.

Deviations and squared Deviations of the data
xi d=xi-x- d2
12.01 12.01-12=0.01 0.0001
12.02 12.02-12=0.02 0.0004
11.95 11.95-12=-0.05 0.0025
11.99 11.99-12=-0.01 0.0001
11.94 11.94-12=-0.06 0.0036l
12.01 12.01-12=0.01 0.0001
12.03 12.03-12=0.03 0.0009
11.98 11.98-12=-0.02 0.0004
12.00 12.00-12=0 0
12.03 12.03-12=0.03 0.0009
11.98 11.98-12=-0.02 0.0004
12.05 12.05-12=0.05 0.0025
11.93 11.93-12=-0.07 0.0049
11.98 11.98-12=-0.02 0.0004

(xi-x¯)2=0.0172

Thus, we have

(xix¯)2=0.0172 (Squared deviation)

n=14 (Sample size)

Substituting the above values in

s2=xi-x-2n-1

     =0.017214-1

     =0.017213

 s2=0.0013

Now, construct a confidence interval for the population variance with

n=14 (Sample size)

s2=0.0013 (Sample variance)

c=0.95 (Level of confidence)

Here, we need to construct the confidence interval for the population variance, thus the point estimate is the value of s2 and is given as

s2=0.0013

Calculate α2 and 1-α2, based on the level of confidence given, for the level of confidence given,

c=0.95

We have,

α=1-c

α=1-0.95

α=0.05

Thus,

   =α2

   =0.052

   =0.025 and

1-α2

=1-0.025

=0.975

Find the critical value χα22 and χ1-α22 for the distribution with n-1 degrees of freedom using the χ2-distribution table,

=χα22

=χ0.0252

= 24.74 and

χ1-α22

=χ1-0.0252

=χ0.9752

=5.009

Find the confidence interval by substituting the necessary values in the formula:

n-1s2χα22<σ2<n-1s2χ1-α22

14-10.001324.74<σ2<14-10.00135.009

130.001324.74<σ2<130.00135.009

0.016924.74<σ2<0.01695.009

0.00068<σ2<0.00337

0.001<σ2<0.003

Using, interval notation, the confidence interval can also be written as (0.001, 0.003).

Final statement:

Therefore, the confidence interval for the population variance is given by (0.001, 0.003).

The volumes of soda in all the soda cans that come off that particular assumbly line will have the variance between 0.001 and 0.003.

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Chapter 8 Solutions

Beginning Statistics, 2nd Edition

Ch. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.2 - Prob. 1ECh. 8.2 - Prob. 2ECh. 8.2 - Prob. 3ECh. 8.2 - Prob. 4ECh. 8.2 - Prob. 5ECh. 8.2 - Prob. 6ECh. 8.2 - Prob. 7ECh. 8.2 - Prob. 8ECh. 8.2 - Prob. 9ECh. 8.2 - Prob. 10ECh. 8.2 - Prob. 11ECh. 8.2 - Prob. 12ECh. 8.2 - Prob. 13ECh. 8.2 - Prob. 14ECh. 8.2 - Prob. 15ECh. 8.2 - Prob. 16ECh. 8.2 - Prob. 17ECh. 8.2 - Prob. 18ECh. 8.2 - Prob. 19ECh. 8.2 - Prob. 20ECh. 8.2 - Prob. 21ECh. 8.2 - Prob. 22ECh. 8.2 - Prob. 23ECh. 8.2 - Prob. 24ECh. 8.2 - Prob. 25ECh. 8.2 - Prob. 26ECh. 8.3 - Prob. 1ECh. 8.3 - Prob. 2ECh. 8.3 - Prob. 3ECh. 8.3 - Prob. 4ECh. 8.3 - Prob. 5ECh. 8.3 - Prob. 6ECh. 8.3 - Prob. 7ECh. 8.3 - Prob. 8ECh. 8.3 - Prob. 9ECh. 8.3 - Prob. 10ECh. 8.3 - Prob. 11ECh. 8.3 - Prob. 12ECh. 8.3 - Prob. 13ECh. 8.3 - Prob. 14ECh. 8.3 - Prob. 15ECh. 8.3 - Prob. 16ECh. 8.3 - Prob. 17ECh. 8.3 - Prob. 18ECh. 8.3 - Prob. 19ECh. 8.3 - Prob. 20ECh. 8.3 - Prob. 21ECh. 8.3 - Prob. 22ECh. 8.3 - Prob. 23ECh. 8.3 - Prob. 24ECh. 8.3 - Prob. 25ECh. 8.3 - Prob. 26ECh. 8.3 - Prob. 27ECh. 8.4 - Prob. 1ECh. 8.4 - Prob. 2ECh. 8.4 - Prob. 3ECh. 8.4 - Prob. 4ECh. 8.4 - Prob. 5ECh. 8.4 - Prob. 6ECh. 8.4 - Prob. 7ECh. 8.4 - Prob. 8ECh. 8.4 - Prob. 9ECh. 8.4 - Prob. 10ECh. 8.4 - Prob. 11ECh. 8.4 - Prob. 12ECh. 8.4 - Prob. 13ECh. 8.4 - Prob. 14ECh. 8.4 - Prob. 15ECh. 8.4 - Prob. 16ECh. 8.4 - Prob. 17ECh. 8.4 - Prob. 18ECh. 8.4 - Prob. 19ECh. 8.4 - Prob. 20ECh. 8.4 - Prob. 21ECh. 8.5 - Prob. 1ECh. 8.5 - Prob. 2ECh. 8.5 - Prob. 3ECh. 8.5 - Prob. 4ECh. 8.5 - Prob. 5ECh. 8.5 - Prob. 6ECh. 8.5 - Prob. 7ECh. 8.5 - Prob. 8ECh. 8.5 - Prob. 9ECh. 8.5 - Prob. 10ECh. 8.5 - Prob. 11ECh. 8.5 - Prob. 12ECh. 8.5 - Prob. 13ECh. 8.5 - Prob. 14ECh. 8.5 - Prob. 15ECh. 8.5 - Prob. 16ECh. 8.5 - Prob. 17ECh. 8.5 - Prob. 18ECh. 8.5 - Prob. 19ECh. 8.5 - Prob. 20ECh. 8.5 - Prob. 21ECh. 8.5 - Prob. 22ECh. 8.5 - Prob. 23ECh. 8.5 - Prob. 24ECh. 8.5 - Prob. 25ECh. 8.5 - Prob. 26ECh. 8.5 - Prob. 27ECh. 8.5 - Prob. 28ECh. 8.5 - Prob. 29ECh. 8.5 - Prob. 30ECh. 8.CR - Prob. 1CRCh. 8.CR - Prob. 2CRCh. 8.CR - Prob. 3CRCh. 8.CR - Prob. 4CRCh. 8.CR - Prob. 5CRCh. 8.CR - Prob. 6CRCh. 8.CR - Prob. 7CRCh. 8.CR - Prob. 8CRCh. 8.CR - Prob. 9CRCh. 8.CR - Prob. 10CRCh. 8.CR - Prob. 11CRCh. 8.CR - Prob. 12CRCh. 8.CR - Prob. 13CRCh. 8.PA - Prob. 1PCh. 8.PA - Prob. 2PCh. 8.PA - Prob. 3PCh. 8.PA - Prob. 4PCh. 8.PA - Prob. 5PCh. 8.PB - Prob. 1PCh. 8.PB - Prob. 2PCh. 8.PB - Prob. 3PCh. 8.PB - Prob. 4PCh. 8.PB - Prob. 5P
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