Workers at a private business was given a grade (A, B, C) based on their yearly performance. The data was recorded along with the department they belong. The results obtained using MINITAB are shown below: Expected counts are printed below observed counts A B C Total 12 8 7 Accounting (12.960) (7.830) (6.210) * 17 10 8 Sales (16.800) (10.150) (8.050) 35 19 11 8 Marketing (18.240) (**) (8.740) 38 Total 48 29 23 100 Chi-sq = 0.0711 + 0.0037 + 0.1005 + 0.0024 + 0.0022 + 0.003 + 0.0317 + 0.0000 + *** = 0.2746 Degrees of Freedom = ??, p-value = ??? What is the conclusion of this test? Give reasons for your answer
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
Workers at a private business was given a grade (A, B, C) based on their yearly performance. The data was recorded along with the department they belong. The results obtained using MINITAB are shown below:
Expected counts are printed below observed counts
A B C Total
12 8 7
Accounting (12.960) (7.830) (6.210) *
17 10 8
Sales (16.800) (10.150) (8.050) 35
19 11 8
Marketing (18.240) (**) (8.740) 38
Total 48 29 23 100
Chi-sq = 0.0711 + 0.0037 + 0.1005 +
0.0024 + 0.0022 + 0.003 +
0.0317 + 0.0000 + *** = 0.2746
Degrees of Freedom = ??, p-value = ???
What is the conclusion of this test? Give reasons for your answer
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