Use the given degree of confidence and sample data to construct a confidence interval for the population mean?

Use the given degree of confidence and sample data to construct a confidence interval for the population mean 渭. Assume that the population has a normal distribution.

28) A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers. 28) ______



A) $453.59 %26lt; 渭 %26lt; $874.69

B) $455.65 %26lt; 渭 %26lt; $872.63

C) $493.71 %26lt; 渭 %26lt; $834.57

D) $492.52 %26lt; 渭 %26lt; $835.76





29) The principal randomly selected six students to take an aptitude test. Their scores were:

83.0 84.1 83.5 83.7 84.1 73.5



Determine a 90 percent confidence interval for the mean score for all students. 29) ______



A) 78.45 %26lt; 渭 %26lt; 85.52

B) 85.42 %26lt; 渭 %26lt; 78.55

C) 85.52 %26lt; 渭 %26lt; 78.45

D) 78.55 %26lt; 渭 %26lt; 85.42





30) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:



7.6 10.4 9.7 8.4 11.8

7.0 6.5 11.1 10.4 12.4

Determine a 95 percent confidence interval for the mean time for all players. 30) ______

A) 11.03 %26lt; 渭 %26lt; 8.03

B) 8.13 %26lt; 渭 %26lt; 10.93

C) 8.03 %26lt; 渭 %26lt; 11.03

D) 10.93 %26lt; 渭 %26lt; 8.13Use the given degree of confidence and sample data to construct a confidence interval for the population mean?
28) ANSWER: A) $453.59 %26lt; 渭 %26lt; $874.69



Why???



SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION

x-bar = Sample mean 664.14

s = Sample standard deviation297.29

n = Number of samples 14

df = degrees of freedom 13



significant digits2



Confidence Level98

"Look-up" Table 't-critical value'2.65

Look-up Table of t critical values for confidence and prediction intervals. Central two-side area = 98%

with df = 13. Another Look-up method is to utilize Microsoft Excel function:

TINV(probability,degrees_freedom) Returns the inverse of the Student's t-distribution



98% Resulting Confidence Interval for 'true mean':

x-bar +/- ('t critical value') * s/SQRT(n)= 664.14 +/- 2.65 * 297.29/SQRT(14) = [453.59, 874.69]









29) ANSWER: D) 78.55 %26lt; 渭 %26lt; 85.42



90% Resulting Confidence Interval for 'true mean':

x-bar +/- ('t critical value') * s/SQRT(n)= 81.98 +/- 2.02 * 4.18/SQRT(6) = [78.55, 85.42]











30) ANSWER: C) 8.03 %26lt; 渭 %26lt; 11.03



95% Resulting Confidence Interval for 'true mean':

x-bar +/- ('t critical value') * s/SQRT(n)= 9.52 +/- 2.26 * 2.05/SQRT(10) = [8.03, 11.03]