solutions hw 6

solutions hw 6 - 1 Which of the fol ow ing statem ents...

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Unformatted text preview: 1) Which of the fol ow ing statem ents about Type I and Type I er ors is cor ect a) Type I: R eject a true nul hypothesis. Type I : D o not reject a false nul hypothesis. 2) In a test of hypothesis, the sym bol α represents: c) The level of significance, or the probability of com m it ing a Type I er or. 3) d ) Probability value = 0.032 < α = .05 Reject H ₀ Probability value = 0.032 > α = .01 Do not eject H ₀ 4) d ) 0.0 75 H : μ ≤ ₀ 50 0 H : μ > ₁ 50 0 Up er tail test |z | = 2.43 probability value = P(z > 2.43) = 0.0 75 5) c) H : μ ≤ 40 ₀ H : μ > 40 ₁ Your are testing the hypothesis that the average age is over (greater than) 40. The statement "> 40" must ap ear as the alternative hypothesis. 6) In the previous question, b ) |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 42 s = 10 n = 219 0.676 |z| = 2.96 α = 0.05 1.64 7) C onsider the fol ow ing hypothesis test. a) |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 9.46 s = 2 n = 50 0.283 |z| = 1.91 α = 0.05 1.64 8) What is the probability value of the test statistic? b ) 0.0281 |z | = 1.91 probability value = P(z > 1.91) = 0.0281 9) C onsider the fol ow ing hypothesis test. b ) |z | = 1.25. D o not reject H 0. |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 26.5 s = 12 n = 10 1.20 |z| = 1.25 α = 0.05 1.64 10) 6.7 6.7 b ) |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 7.25 s = 2.5 n = 20 0.17 |z| = 3.1 α = 0.01 2.3 1 ) H 0: μ ≥ 10,192 H 1: μ < 10,192 a) |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 930 s = 450 n = 10 450 |z| = 1.98 α = 0.05 P(z > |z|) = 0.0239 Reject H if probability value < α ₀ 12) The hypotheses for this test are: b ) 15 15 We are testing the nul hypothesis that the mean is 15 minutes or les (les than or equal to 15) against the alternative that the mean is greater than 15. 13) U sing a 1% level of significance, b ) prob-value = 0.0 15. The prem ium rate should be charged. |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 17 s = 4 n = 35 0.676 |z| = 2.96 α = 0.05 P(z > |z|) = 0.0 15 14) c) 37,0 0 37,0 0 15) In the previous question, using a 5% level of significance, a) prob-value = 0.0708. C an ot conclude population m ean salary has increased in June 20 1. |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 3810 s = 520 n = 48 750.56 |z| = 1.47 α = 0.05 P(z > |z|) = 0.0708 16) a) prob value is 0.0873. C an ot conclude that the population m ean is dif erent from 8. 8 8 |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 7.5 s = 3.2 n = 120 0.292 |z| = 1.71 α = 0.05 P(z > |z|) = 0.0436 Reject H if probability value < α ₀ prob value = 2 × 0.0436 = 0.0873 Do not reject the nul hypothesis. Can ot conclude that the mean is dif erent from 8 minutes. 17) c) |z | = 2.54. C onclude the m ean is dif erent from 16. Shut the line dow n. 16 16 |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 16.32 s = 0.8 n = 40 0.126 |z| = 2.54 α = 0.05 1.64 18) c) 90 90 |z| = (x − μ ) ∕ se(x) ̄ ₀ ̄ 935 s = 180 n = 20 12.728 |z| = 2.75 α = 0.05 P(z > |z|) = 0.0 30 prob value = 2 × 0.0 3 = 0.0 60 Reject H if probability value < α ₀ 24.95 |x − μ |=...
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This document was uploaded on 10/27/2011 for the course ECON-E 270 at Indiana.

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