Question

Use the given degree of confidence and the sample data to construct a confidence interval for the population mean u. Assuming the population has a normal distribution. The Administrator randomly selected six employees to take an aptitude test. Their scores were: 9.3 81.2 74.9 83.2 83.7 85.5 Determine the margin of error for the 90% confidence interval for the average score for all employees.

Answer 1

solution:-

given that n = 6

mean = 69.63 , standard deviation = 29.7836

degree of freedom df = n - 1 = 6 - 1 = 5

we look into t table with df and (1-0.90) = 0.10 probability

critical value t = 2.015

=> margin of error formula

= t * standard deviation/sqrt(n)

= 2.015 * 29.7836/sqrt(6)

= 24.5006 (rounded to four decimal places)

=> confidence interval formula

= mean +/- margin of error

= 69.63 +/- 24.5006

= (45.1294 , 94.1306) rounded to four decimal places