Question

In: Statistics and Probability

You are to test if the average SAT score in the high school students of Ontario...

You are to test if the average SAT score in the high school students of Ontario is greater than 495. You set the hypotheses as below:

H0: µ = 495 vs Ha: µ > 495

If the SRS of 500 students were selected and the standard deviation has been known to 100, with alpha = 0.05, answer the followings.

a.  What is the Type I error?

b. What is the Type II error if the true population mean is 510. Explain it.

c. What is the power if the true population mean is 510? Explain it.

Please show your work and thank you SO much in advance! You are helping a struggling stats student SO much!

Solutions

Expert Solution

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
Critical value Z(α/2) = Z( 0.05 /2 ) = 1.645
1.645 = ( X̅ - 495 ) / ( 100 / √( 500 ))
X̅ >= 502.3567

Part a)

X ~ N ( µ = 495 , σ = 100 )
P ( X > 502.3567 ) = 1 - P ( X < 502.3567 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 502.3567 - 495 ) / ( 100 / √ ( 500 ) )
Z = 1.645
P ( ( X - µ ) / ( σ / √ (n)) > ( 502.3567 - 495 ) / ( 100 / √(500) )
P ( Z > 1.64 )
P ( X̅ > 502.3567 ) = 1 - P ( Z < 1.64 )
P ( X̅ > 502.3567 ) = 1 - 0.95
P ( X̅ > 502.3567 ) = 0.05
P ( Type I error ) α = 0.05

Part b)

X ~ N ( µ = 510 , σ = 100 )
P ( X < 502.3567 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 502.3567 - 510 ) / ( 100 / √500 )
Z = -1.7091
P ( ( X - µ ) / ( σ/√(n)) < ( 502.3567 - 510 ) / ( 100 / √(500) )
P ( X < 502.3567 ) = P ( Z < -1.71 )
P ( X̅ < 502.3567 ) = 0.0437
P ( X̅1 < 502.3567 | µ = 510 ) = 0.0437
P ( Type II error ) ß = 0.0437

Part c)

Power of test is ( 1 - ß ) = 0.9563

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