CONTINGENCY TABLES 250 0.2113 0.2925 0.8403 0.5166 0.3900 0.3768 0.4993 0.7297 0.4632 0.4403 0.7979 0.8757 0.8312 0.5590 0.8522 0.9763 0.8230 0.4295 0.5836 0.4004 0.3003 0.8432 0.3003 h 0.2595 0.9329 0.2172 0.4396 0.4887 0.8008 0.4579 0.5274 0.5337 0.4067 0.3686 0.6793 0.5498 0.2172 0.4995 0.9096 A die was cast 600 times with the following results. 4 3 2. CO 5 6 2 1 Оссиrrence 122 89 98 108 best 87 96 Frequency 3. Use the number of hits for each player in Example 2 to test the null hypothesis that all ts distribution is that the probability is the same for all trials, and this is one way of testing Is the die balanced? players have the same probability of getting a hit. Note that one assumption of the binomial that assumption in Example 2. Without the aid of books or tables, attempt to write 300 random digits. Then apply bhe test of randomness described in Example 2 to see if you are a good random digit generator The number of babies born in Methodist Hospital last year was as follows. In the Winter there were 36 babies born, in the Spring 45, in the Summer 42, and in the Fall 55. Test the hypothesis that the number of births is uniformly distributed over the four seasons of 4. 5. the year. Twenty-six observations were obtained, and the question arose as to whether they followed a normal distribution with mean 12 and standard deviation 3. None of the observations were below the lower quartile of this distribution, and 12 were above the upper quartile. Six were below the median, and 8 were between the median and the upper quartile. D these observations appear to have come from the distribution described? 6. bo 4.6 COCHRAN'S TEST FOR RELATED OBSERVATIONS Sometimes the use of a treatment, or condition, results in one of two possible outcomes. For example, the response to a salesperson's technique may be classifted as "sale" or "no sale," or a certain treatment may result in "success" or "failure." Of course, if several treatments, c in number, are each applied in several differenm independent trials, the results may be given in the form of a 2 x c contingeny table, where one row represents the number of successes and the other row represents the number of failures, and the null hypothesis of no treatment differ ences may be tested using a chi-squared contingency table test, as described it Section 4.2. However, it is often possibl treatments, that is, increase the po t more subtle differences between est, by applying all c treatments