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Unformatted text preview: 3. Suppose n = 49 observations are taken from a normal distribution where σ = 8.0 for the purpose of testing H 0 : µ = 60 versus H 1 : µ ≠ 60 at the α = 0.05 level of significance. a) What is the power of the appropriate test when µ = 59.2? H 0 : µ = 60 vs. H 1 : µ ≠ 60
Rejection Region: Reject H 0 if Z=
X − µ0 2 – tailed. α = 0.05. σ < − zα 2 or Z= X − µ0 n σ > zα 2 n σ n
8 49 ⇒ ⇒ ⇒ X < µ 0 − zα 2 X < 60 − 1.96 σ n
8 49 or or or X > µ 0 + zα 2 X > 60 + 1.96 X < 57.76 X > 62.24 Power = P( Reject H 0  H 0 is false ) = P( X < 57.76  µ = 59.2 ) + P( X > 62.24  µ = 59.2 ) 57.76 − 59.2 62.24 − 59.2 = P Z < + P Z > 8 8 49 49 = P( Z < – 1.26 ) + P( Z > 2.66 ) = 0.1038 + 0.0039 = 0.1077. b) What is the pvalue of the test if the observed value of the sample mean is x = 58?
58 − 60 = – 1.75. 8 49 The observed value of the test statistic is
2 – tailed test. z= Pvalue = 2 × P ( Z < – 1.75 ) = 2 × 0.0401 = 0.0802. 4. At Anytown State University, the population standard deviation of the SAT scores of entering students is 72. Last year’s overall mean SAT score of entering students was 1100. We suspect that this year’s overall mean SAT score is higher. Find the power of the appropriate test at µ = 1111, if the decision is be based on a random sample of 81 students from this year's class, and a 5% level of significance is used. Let µ = this year’s overall mean SAT score. Claim: µ > 1100. H 0 : µ ≤ 1100 vs. H 1 : µ > 1100.
Rejection Region: Reject H 0 if Right – tailed. Z= X − µ0 σ n
⇒ > zα ⇒ Z=
72 81 X − 1100 > 1.645 72 81 X > 1100 + 1.645 ⋅ = 1113.16. Power = P ( Reject H 0  H 0 is NOT true ) 1113.16 − 1111 = P ( X > 1113.16  µ = 1111 ) = P Z > = P ( Z > 0.27 ) 72 81 = 1 – Φ ( 0.27 ) = 1 – 0.6064 = 0.3936. ...
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This note was uploaded on 10/29/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Summer '08 term at University of Illinois at Urbana–Champaign.
 Summer '08
 AlexeiStepanov
 Normal Distribution

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