Question

Students collected data measuring the accuracy of wristwatches. The times (sec) below show the discrepancy between the real time and the time indicated on the wristwatch. Negative values correspond to watches that are running ahead of the actual time. The data appears to be from a normally distributed population. Construct a 95% confidence interval estimate of the mean discrepancy of the population of wristwatches. -85 325 20 305 -93 15 282 27 555 570 -241 36

Answer 1

solution:-

given values = -85 ,325, 20, 305 ,-93 ,15 ,282, 27, 555, 570, -241 ,36

mean = 143

standard deviation = 259.7754

n = 12

degree of freedom df = n - 1 = 12 - 1 = 11

we look into t table with df and two tail probability of (1-0.95) = 0.05

critical value t = 2.201

confidence interval formula

=> mean +/- t * standard deviation/sqrt(n)

=> 143 +/- 2.201 * 259.7754/sqrt(12)

=> (-22.0545 , 308.0545)

=> (-22.1 , 308.1) rounded to one decimal place