11_02_11_1ans1 - 2 α χ Reject H b Use α = 0.05 to perform a large-sample test H p = 0.60 vs H 1 p ≠ 0.60 200 105 ˆ = p = 0.525 Z = 200 40 60

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STAT 409 Examples for 11/02/2011 (1) Fall 2011 1. 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 = ( ) ( ) 80 80 95 120 120 105 2 2 - + - = 1.875 + 2.8125 = 4.6875 . 2 – 1 = 1 degree of freedom. 2 05 . 0 χ ( 1 ) = 3.841 . Q 1 = 4.6875 > 3.841 =
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Unformatted text preview: 2 α χ . Reject H . b) Use α = 0.05 to perform a large-sample test H : p = 0.60 vs. H 1 : p ≠ 0.60. 200 105 ˆ = p = 0.525. Z = 200 40 . 60 . 60 . 525 . ⋅-= – 2.165 . ± z 0.025 = ± 1.960. Z = – 2.165 < – 1.960 = – z α / 2 . Reject H . c) Compare the test statistics Q 1 ( part (a) ) and Z ( part (b) ). Compare the critical values 2 05 . χ ( part (a) ) and ± z 0.025 ( part (b) ). Q 1 ( part (a) ) = [ Z ( part (b) ) ] 2 2 05 . χ ( part (a) ) = [ ± z 0.025 ( part (b) ) ] 2...
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This note was uploaded on 11/15/2011 for the course STAT 409 taught by Professor Stephanov during the Fall '11 term at University of Illinois at Urbana–Champaign.

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