Conducting your own Chi-square 4 different kinds of beans have been mixed and pl

Question

Conducting your own Chi-square 4 different kinds of beans have been mixed and placed into a large container. The beans should all exist in equal amounts or a ratio of 1:1:1:1. Remove a random sample of the bean mixture. Separate and count each bean type. Fill in the table below Calculation of ?2 for a sample removed from a population Type of Observed Expected (O-E)(O-E)(O-E)/E Bean 0.06 Totals 311 D63 2.2 1) What is our null hypothesis (in terms of bean sampling) for this test? 2) What is our alternative hypothesis? 3) How many degrees of freedom are there? hat probability does your x value correspond to? P 0. 004 0 63 5) s this value significant or not significant (according to the x table)? 6) Which hypothesis is supported by the results of your X test? 7) Please write your interpretation of the results of this experiment (what do your results tell you about your sampled population of beans)? After the P values for each class member have been tabulated on the board, answer the following question. 8) Why do the P values vary within the class when the actual ratio of beans is 1:1:1:1?

Explanation / Answer

1. The null hypothesis is that there is no difference between the expected and observed values, and that the sampling is random

2. The alternate hypothesis is that there exists significant differenc between the expected and the observed values, and that the sampling is not random

3. Degrees of freedom = 4-1 = 3

4. The P value = 0.0041

5. The P value is below 0.05, and hence, is significant

6. Alternate hypothesis, since P value is below 0.05, null hypothesis is rejected

P value = 0.004062986

7. The beans are not sampled randomly.

Obs Exp O-E (O-E)2 (O-E)2/E White 94 77.75 16.25 264.063 3.40 Black 88 77.75 10.25 105.063 1.35 Red 77 77.75 -0.75 0.5625 0.01 Brown 52 77.75 -25.75 663.063 8.53 Totals 311 311 13.28

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