Chapter 26final.pdf

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Biology (Biological Sciences)
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Blaine Mullins

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Chapter 26: Chi – Square Tests In this chapter, we will talk about:  A goodness – of – fit test,  Chi-square test of independence in a contingency table.  Chi-square test of homogeneity in a contingency table. Definition: A goodness-of-fit test is one that is used to see if the distribution of the observed outcome of the sample trials supports a hypothesized population distribution. Note: We cannot prove a null hypothesis using a statistical test. One of the tests is k (ObExp) 2 2 TS   Exp ~  distribution with df  k 1 i1 Example 26.1: A die was rolled 600 times, with the accompanying results. Do these data present sufficient evid ence to indicate that the die is unbalanced? Observed number 1 2 3 4 5 6 Frequency 89113 98 104 11779 Solution: H 0 1) H : a 2)   0.05 k (ObsEp) 2 2 3) TS  ~  distribution Hf0is true. i1 Exp df  k 1 4) We reject H 0ifTS  2* 5) 6) Since, At   0.05, there enough evidence to conclude that the die is unbalanced. 7) p  value  Example 26.2: The Mendelian theory states that the number of a certain type of peas falling into the classifications round and yellow, wrinkled and yellow, round and green, and wrinkled and green should be in the ratio 9:3:3:1. Suppose that 100 such peas revealed 56, 19, 17, and 8 in the respective classes. Are these data consistent with the model? Use   0.05 (Hint: The expression 9:3:3:1 means that 9 of the peas should be round 16 3 and yellow, should be wrinkled and yellow, etc.) 16 Example 26.3: The editor of a newspaper has written that 25 percent of the university students, in the paper's circulation area, read newspapers daily. A random sample of 200 of these university students shows that 45 of them are daily readers of newspapers. At   0.025 level, is the editor's statement likely to be true? Solution: Hypothesis test results: p : proportion of successes for population H 0 p = 0.25 H A p ≠ 0.25 Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 45 200 0.225 0.030618621 -0.8164966 0.4142 Chi-Square goodness-of-fit results: Observed: Obs. Expected: Exp. N DF Chi-Square P-Value 200 1 0.6666667 0.4142 Example 26.4: Applicants for public assist ance are allowed an appeals process when they feel unfairly treated. At such a hearing, the applicant may choose self-representation or represen tation by an attorney. The appeal may result in an increase, decrease, or no change of the aid recommendation. Court records of 320 appeals cases provided the following data: Amount of Aid Type of Representation Increased Unchanged Decreased Total Self 59 108 17 184 Attorney 70 63 3 136 Total 129 171 20 320 Test the null hypothesis that the appeals decision and the type of representation are independent. Solution: H0:The appeals decision and the type of representation are independent. 1) H :The appeals decision and the type of representation are not independent. a 2)   0.05 k (ObExp) 2 3) TS   ~   distributionHi0is true. i1 Exp df)()(1  C 4) We reject H if TS  2* 0 The expected values are calculated from the following formula: Expected value =Row Total  Column Total Grand Total 5) Type of Representation Increased Unchanged Decreased Total 184 129 184171 184 20 Self 320 74.175 320 98.325 320 11.5
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