Degrees of freedom df = n-1 = 20 - 1 = 19
99% confidence interval for
is
Sqrt [ (n-1) S2 / 
/2 ] <
< Sqrt [
(n-1) S2 /
1-
/2 ]
Sqrt [ 19 * 2.72 / 38.582 ] <
< sqrt [
19 * 2.72 / 6.844 ]
1.9 <
< 4.5
99% CI for
is ( 1.9
, 4.5 )
Use the given degree of confidence and sample data to find a confidence interval for the...
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal pl aces as the sample standard deviation. 21) A sociologist develops a test to measure attitudes about public transportation, and 27 21) randomly selected subjects are given the test. Their mean scoreis 76.2 and their standard deviation is 21.4. Construct the...
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. 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. College students' annual earnings: 98% confidence; n = 9, sample mean is $4121, s = $873 Group of answer choices: a) $551 < σ < $1925 b) $590 < σ < $1672 c) $531 <...
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . 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. 21) A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95%...
(1 poi 16. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation o. 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. Weights of men: 90% confidence; n = 14, x = 161.0 lb, s = 10.9 lb 8.1 lb<o< 15.3 lb 8.3 lb < < 16.2 lb 8.8 lb <0 < 2.7 lb...
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation o. 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. College students' annual earnings: 98% confidence; n = 9, sample mean is $4121, s = $873 $551<o< $1925 O $590 < 0 < $1672 $531 < o < $1708 $687 <o< $1139
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. 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. The amounts (in ounces) of juice in eight randomly selected juice bottles are: 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 σ. 0.22...
Use the given degree of confidence and sample data to construct a confidence interval for the population mean p. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 76 and the standard deviation was 7.7, construct a 99% confidence interval for the mean score of all students. 72.13 < x < 79.87 73.61 < p < 78.39 O 72.54 < p < 79.46 O 72.14< p < 79.89
Use the given degree of confidence and sample data to construct a confidence interval for the population mean H. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 82 and the standard deviation was 12.6, construct a 99% confidence interval for the mean score of all students. Round to two decimal places. O A. 75.68 < < 88.32 OB. 75.66<u < 88.34 OC. 78.09 < < 85.91 OD....
Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 75 and the standard deviation was 5.8, construct a 99% confidence interval for the mean score of all students. Round to two decimal places. A. 73.20<μ<76.80 B. 72.09<μ<77.91 C. 72.08<μ<77.92 D. 72.39<μ<77.61
inutes, The duration of telephone calls directed by the local telephone company: s 3.8 m 100, confidence interval 98%. 4. The mean replacement time for a random sample of 20 machines is 7.5 years and the standard deviation is 2.4 years. Construct a 99% confidence interval for the standard deviation , σ , of a replacement time of all washing machines of this type (user distribution for analysis)