Question

multiple choice. choose the one alternative that best completes the statement or answers the question.
solve the problem.

1. find the critical value ( chi_{r}^{2} ) corresponding to a sample size of 19 and a confidence level of 99

1)
percent.
a) 37.156
b) 7.015
c) 34.805
d) 6.265

1. find the chi - square value ( chi_{l}^{2} ) corresponding to a sample size of 4 and a confidence level of 98

2)
percent.
a) 11.345
b) 0.115
c) 9.348
d) 0.216
use the given degree of confidence and sample data to find a confidence interval for the population standard
deviation ( sigma ). assume that the population has a normal distribution. round the confidence interval limits to the same
number of decimal places as the sample standard deviation.

1. weights of eggs: 95% confidence; ( n = 22, overline{x}=1.65 mathrm{oz}, s = 0.47 mathrm{oz} )

3)
a) ( 0.36 mathrm{oz}<sigma<0.65 mathrm{oz} )
b) ( 0.38 mathrm{oz}<sigma<0.63 mathrm{oz} )
c) ( 0.36 mathrm{oz}<sigma<0.67 mathrm{oz} )
d) ( 0.37 mathrm{oz}<sigma<0.61 mathrm{oz} )

1. the mean replacement time for a random sample of 20 washing machines is 10.9 years and the

4)
standard deviation is 2.7 years. construct a 99% confidence interval for the standard deviation,
( sigma ), of the replacement times of all washing machines of this type.
a) ( 1.9 mathrm{yr}<sigma<5.7 mathrm{yr} )
b) ( 2.0 mathrm{yr}<sigma<4.3 mathrm{yr} )
c) ( 1.9 mathrm{yr}<sigma<4.5 mathrm{yr} )
d) ( 1.8 mathrm{yr}<sigma<5.1 mathrm{yr} )

1. a sociologist develops a test to measure attitudes about public transportation, and 27 randomly

5)
selected subjects are given the test. their mean score is 76.2 and their standard deviation is 21.4.
construct the 95% confidence interval for the standard deviation, ( sigma ), of the scores of all subjects.
a) ( 16.6<sigma<28.6 )
b) ( 17.2<sigma<27.2 )
c) ( 16.9<sigma<29.3 )
d) ( 17.5<sigma<27.8 )
use the given degree of confidence and sample data to find a confidence interval for the population standard
deviation ( sigma ). assume that the population has a normal distribution. round the confidence interval limits to one more
decimal place than is used for the original set of data.

1. the amounts (in ounces) of juice in eight randomly selected juice bottles are:

6)
15.2 15.1 15.9 15.5
15.6 15.1 15.8 15.0
find a 98% confidence interval for the population standard deviation ( sigma ).
a) ( 0.21 mathrm{oz}<sigma<0.82 mathrm{oz} )
b) ( 0.21 mathrm{oz}<sigma<0.71 mathrm{oz} )
c) ( 0.20 mathrm{oz}<sigma<0.71 mathrm{oz} )
d) ( 0.22 mathrm{oz}<sigma<0.86 mathrm{oz} )

Explanation:

Step1: Determine the degrees of freedom

For a chi - square distribution, the degrees of freedom \(df=n - 1\).

For problem 1:

Given \(n = 19\), then \(df=19 - 1=18\).
For a 99% confidence level, the right - tailed critical value \(\chi_{R}^{2}\) is used. Looking up in the chi - square distribution table, \(\chi_{0.005,18}^{2}=37.156\) (since \(\alpha=(1 - 0.99)/2 = 0.005\) for the right - tail in a two - tailed confidence interval for variance/standard deviation).

For problem 2:

Given \(n = 4\), then \(df=4 - 1=3\).
For a 98% confidence level, the left - tailed critical value \(\chi_{L}^{2}\) is used. Since \(\alpha=(1 - 0.98)/2=0.01\) for the left - tail in a two - tailed confidence interval for variance/standard deviation, looking up in the chi - square distribution table, \(\chi_{0.99,3}^{2}=0.115\) (\(\chi_{1-\alpha,df}^{2}\) for the left - tail).

Step2: Calculate the confidence interval for standard deviation

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{U}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\)

For problem 3:

Given \(n = 22\), \(s = 0.47\), \(df=22-1 = 21\), for 95% confidence level, \(\chi_{0.025,21}^{2}=35.479\), \(\chi_{0.975,21}^{2}=10.283\)
\(\sqrt{\frac{(22 - 1)\times(0.47)^{2}}{35.479}}<\sigma<\sqrt{\frac{(22 - 1)\times(0.47)^{2}}{10.283}}\)
\(\sqrt{\frac{21\times0.2209}{35.479}}<\sigma<\sqrt{\frac{21\times0.2209}{10.283}}\)
\(\sqrt{\frac{4.6389}{35.479}}<\sigma<\sqrt{\frac{4.6389}{10.283}}\)
\(\sqrt{0.1308}<\sigma<\sqrt{0.4511}\)
\(0.36<\sigma<0.67\)

For problem 4:

Given \(n = 20\), \(s = 2.7\), \(df=20 - 1=19\), for 99% confidence level, \(\chi_{0.005,19}^{2}=38.582\), \(\chi_{0.995,19}^{2}=6.844\)
\(\sqrt{\frac{(20 - 1)\times(2.7)^{2}}{38.582}}<\sigma<\sqrt{\frac{(20 - 1)\times(2.7)^{2}}{6.844}}\)
\(\sqrt{\frac{19\times7.29}{38.582}}<\sigma<\sqrt{\frac{19\times7.29}{6.844}}\)
\(\sqrt{\frac{138.51}{38.582}}<\sigma<\sqrt{\frac{138.51}{6.844}}\)
\(\sqrt{3.59}<\sigma<\sqrt{20.24}\)
\(1.9<\sigma<4.5\)

For problem 5:

Given \(n = 27\), \(s = 21.4\), \(df=27-1 = 26\), for 95% confidence level, \(\chi_{0.025,26}^{2}=41.923\), \(\chi_{0.975,26}^{2}=13.844\)
\(\sqrt{\frac{(27 - 1)\times(21.4)^{2}}{41.923}}<\sigma<\sqrt{\frac{(27 - 1)\times(21.4)^{2}}{13.844}}\)
\(\sqrt{\frac{26\times457.96}{41.923}}<\sigma<\sqrt{\frac{26\times457.96}{13.844}}\)
\(\sqrt{\frac{11906.96}{41.923}}<\sigma<\sqrt{\frac{11906.96}{13.844}}\)
\(\sqrt{284}<\sigma<\sqrt{860}\)
\(16.9<\sigma<29.3\)

For problem 6:

First, calculate the sample standard deviation \(s\)
The data set \(x=\{15.2,15.1,15.9,15.5,15.6,15.1,15.8,15.0\}\)
\(\bar{x}=\frac{15.2 + 15.1+15.9+15.5+15.6+15.1+15.8+15.0}{8}=\frac{123.2}{8}=15.4\)
\(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{(15.2 - 15.4)^{2}+(15.1 - 15.4)^{2}+(15.9 - 15.4)^{2}+(15.5 - 15.4)^{2}+(15.6 - 15.4)^{2}+(15.1 - 15.4)^{2}+(15.8 - 15.4)^{2}+(15.0 - 15.4)^{2}}{7}}\)
\(s=\sqrt{\frac{(- 0.2)^{2}+(-0.3)^{2}+(0.5)^{2}+(0.1)^{2}+(0.2)^{2}+(-0.3)^{2}+(0.4)^{2}+(-0.4)^{2}}{7}}=\sqrt{\frac{0.04 + 0.09+0.25+0.01+0.04+0.09+0.16+0.16}{7}}=\sqrt{\frac{0.84}{7}}=\sqrt{0.12}\approx0.35\)
\(n = 8\), \(df=8 - 1=7\), for 98% confidence level, \(\chi_{0.01,7}^{2}=18.475\), \(\chi_{0.99,7}^{2}=1.239\)
\(\sqrt{\frac{(8 - 1)\times(0.35)^{2}}{18.475}}<\sigma<\sqrt{\frac{(8 - 1)\times(0.35)^{2}}{1.239}}\)
\(\sqrt{\frac{7\times0.1225}{18.475}}<\sigma<\sqrt{\frac{7\times0.1225}{1.239}}\)
\(\sqrt{\frac{0.8575}{18.475}}<\sigma<\sqrt{\frac{0.8575}{1.239}}\)
\(\sqrt{0.0464}<\sigma<\sqrt{0.692}\)
\(0…

Answer:

1. A. 37.156
2. B. 0.115
3. C. \(0.36\ oz<\sigma<0.67\ oz\)
4. C. \(1.9\ yr<\sigma<4.5\ yr\)
5. C. \(16.9<\sigma<29.3\)
6. A. \(0.21\ oz<\sigma<0.82\ oz\)