STAT 400 Lecture AL1 Answers for 9.1 Spring 2015 Dalpiaz Pearson’s χ 2 Test for Goodness of Fit ( Based on Large n ) A random sample of size n is classified into k categories or cells. Let Y 1 , Y 2 , … , Y k denote the respective cell frequencies. nkii =∑=1Y. Denote the cell probabilities by p 1 , p 2 , … , p k . H 0 : p 1 = p 10 , p 2 = p 20 , … , p k = p k 0 . 110=∑=kii p. 1 2 … k Total Observed frequency O Y 1 Y 2 … Y k n Probability under H 0 p 10 p 20 … p k 0 1 Expected frequency E under H 0 n p 10 n p 20 … n p k 0 n Test Statistic: ( )( )( )∑∑∑−=−=−===−cells kiiii kiiii kpnpnEEOEEO21210201 YQ Rejection Region: Reject H 0 if Q k – 1 ≥ 2αχ, d.f. = k – 1 = (number of cells) – 1 Pearson’s χ 2 test is an approximate test that is valid only for large samples. As a rule of thumb, n should be large enough so that expected frequency of each cell is at least 5.1. Alex buys a package of Sour Jelly Beans. On the package, it says that 50% of all jelly beans are lemon, 30% are cherry, and 20% are lime. When Alex opens the package, he finds 15 lemon, 9 cherry and 12 lime jelly beans. Is there enough evidence to conclude that the true proportions are different from the ones listed on the package? Use α = 0.05. a) State H 0 and H 1 . H 0 : p 1 = 0.50, p 2 = 0.30, p 3 = 0.20. H 1 : not H 0 . b) Find the expected frequencies under H 0 . n = 15 + 9 + 12 = 36. E 1 = n p 10 = 36 × 0.50 = 18. E 2 = n p 20 = 36 × 0.30 = 10.8. E 3 = n p 30 = 36 × 0.20 = 7.2. c) Calculate the values of the Q test statistic. 1 2 3 Total O 15 9 12 36 E 18 10.8 7.2 36 ( )EEO2 − ( )1818152 − ()10.88.1092 − ( )7.22.7122 − 0.5 0.3 3.2 4.0 d) Find the critical value 2αχ. 3 – 1 = 2 degrees of freedom. 205.0χ( 2 ) = 5.991. e) State your decision ( Accept H 0 or Reject H 0 ) at α = 0.05. Q 2 = 4.0 < 5.991 = 2αχ. Do Not Reject H 0 .2. The financial manager in charge of accounts receivable department is concerned about the current economic slowdown, because customers sometimes wait longer to pay their bills. She wishes to check on this year’s performance of the department by comparing the current outstanding accounts with records from the past few years. Historical records show the following percentages in the respective classifications: Age of Accounts Percent of Receivable Total Receivables Less than 30 days 50% Between 30 and 60 days 25% Between 60 and 90 days 15% Over 90 days 10% To avoid the time required for a complete audit of the accounts receivable, the financial manager chooses a random sample of 60 accounts and finds 24, 18, 15, and 3 accounts, respectively, in the above categories. H 0 : p 1 = 0.50, p 2 = 0.25, p 3 = 0.15, p 4 = 0.10. Perform the appropriate test at the α = 0.05 level of significance. O : 24 18 15 3 E : 60 ⋅ 0.50 = 30 60 ⋅ 0.25 = 15 60 ⋅ 0.15 = 9 60 ⋅ 0.10 = 6 ( )EEO2 −: ( )3030242 − ( )1515182 − ( )99152 − ( )6632 − 1.2 0.6 4.0 1.5 Q = ( )∑−cells EEO2 = 1.2 + 0.6 + 4.0 + 1.5 = 7.3. k – 1 = 4 – 1 = 3 d.f. 205.0 χ = 7.815. Rejection Region: “Reject H 0 if Q ≥ 2αχ” 7.3 = Q < 2αχ = 7.815. Do Not Reject H 0 at α = 0.05. ( 210.0 χ = 6.25. Reject H 0 at α = 0.10. ) ( P-value = 0.062926 )3. The developers of a math proficiency exam to be used at Anytown State University believe that 60% of all incoming freshmen will be able to pass the exam. In a random sample of 200 incoming freshmen, 105 pass the exam. Does this contradict the claim of the developers? a) Use α = 0.05 to perform a goodness-of-fit test H 0 : p 1 = 0.60, p 2 = 0.40 vs. H 1 : H 0 is not true. E 1 = n p 10 = 200 × 0.60 = 120. E 2 = n p 20 = 200 × 0.40 = 80. Q 1 = ( ) ( )80809512012010522−+− = 1.875 + 2.8125 = 4.6875. 2 – 1 = 1 degree of freedom. 205.0χ( 1 ) = 3.841. Q 1 = 4.6875 > 3.841 = 2αχ. Reject H 0 . b) Use α = 0.05 to perform a large-sample test H 0 : p = 0.60 vs. H 1 : p ≠ 0.60. 200105ˆ=p = 0.525. Z = 20040.060.060.0525.0 ⋅− = – 2.165. ± z 0.025 = ± 1.960. Z = – 2.165 < – 1.960 = – z α/2. Reject H 0 . c) Compare the test statistics Q 1 ( part (a) ) and Z ( part (b) ). Compare the critical values 205.0χ ( part (a) ) and ± z 0.025 ( part (b) ). Q 1 ( part (a) ) = [ Z ( part (b) ) ] 2 205.0χ ( part (a) ) = [ ± z 0.025 ( part (b) ) ]
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