Business Statistics help?

10. The manager of a computer software company wishes to study the number of hours senior
executives spend at their desktop computers by type of industry. The manager selected
a sample of five executives from each of three industries. At the .05 significance level,
can she conclude there is a difference in the mean number of hours spent per week by
industry?
Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

Start by noting the descriptive statistics: the mean and std

banking: mean = 10.8; std dev = 1.095
retail: mean = 8; std dev = 1.414
insurance: mean = 8.4; std dev = 1.673

if you knew that the data was normally distributed you could use a student t test to compare two industries. However, because you are looking at three industries you would have to make three different tests. the probability of a type I error increases dramatically in this case. At a 5% significance level you would have a 0.95 probability of making the right conclusion. for three tests that probability of being correct would become 0.95 ^ 3 = 0.857375. One way around this is to use a multiple testing correction like the Bonferroni correction. This is very conservative but to test at the 5% significance level you would run each test at the 0.05 / 3 significance level. if a test had a p-value less than 0.05 / 3 then you have a significant difference at the 0.05 level.

another, an better method would be to use Analysis of Variance. The null hypothesis for the ANOVA is that all the means are equal to each other. the alternate alternative is that one or more of the means is different. If you run an ANOVA you will find a p-value of 0.018 showing that the three industries are not equal in the amount of time that executives spend at their computers.

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