Standard deviation (statistics)

Question

billy has 15 min to get to his after-school job. Despite his best efforts, he is frequently late. His employer says that unless he arrives to work on time consistently, he will lose his job. He has recorded his travel times (in minutes) for the last two weeks: 11, 13, 19, 18, 13, 15, 17, 17, 20, 16. 

over the next two weeks, he continues to record his travel times: 16, 15, 13, 18, 13, 26, 14, 16, 14, 16.

Will Billy lose his job? Use statistics 

Answer 1

This problem is known as a two sample independent t test.

 

The null hypothesis is that the two averages are the same while the alternative hypothesis is that they are different.

 

Average of last two weeks is 15.9 with a standard deviation of 2.884826203 while the average of the next two weeks was 16.1 with a standard deviation of 3.813717929.

 

The t test statistic is -0.1327 and the critical value of t (with df = 18) with a significance level of 0.05 is 2.101.

 

Since 0.1327  is less than 2.101 we cannot conclude that the averages are different and therefore we cannot conclude that his new times are any different than his old times.

 

Hope this helps.

Answer 2

Billy is not consistently one time, obviously, and he's likely to be fired. Perhaps that will teach him a lesson, but it's anybody's guess.  Billy needs to become familiar with the concept of the alarm clock, and probably more than one, and to use it so that on average he gets to work ten minutes early. But it's apparent that Billy is as clueless as can be.