HWsolutions4

# HWsolutions4 - HOMEWORK#4 SOLUTIONS IE121 8-7 = P X > 185...

This preview shows pages 1–2. Sign up to view the full content.

HOMEWORK #4 SOLUTIONS IE121 8-7 a) α = P( X > 185 when μ = 175) = - - 10 / 20 175 185 10 / 20 175 X P = P(Z > 1.58) = 1 - P(Z 1.58) = 1 - 0.94295 = 0.05705 b) β = P( X 185 when μ = 195) = - - 10 / 20 195 185 10 / 20 195 X P = P(Z - 1.58) = 1 - P(Z 1.58) = 0.05705. 8.8 a) Since x >185, we reject the null hypothesis and conclude that the mean foam height is greater than 175 mm. b) P( X > 190 when μ = 175) = P X - - 175 20 10 190 175 20 10 / / = P(Z > 2.37) = 1 - P(Z 2.37) = 1 - 0.99111 = 0.00889. The probability that a value of at least 190 mm would be observed (if the true mean height is 175 mm) is only 0.00889. Thus, the sample value of x = 190 mm would be an unusual result. 8-10 a) ( 29 0571 . 0 175 | = = μ c X P - 16 / 20 175 c Z P = P(Z 1.58) Thus, 1.58 = 16 / 20 175 - c , and c = 182.9 b) If the true mean foam height is 195 mm, then β = P( X 182.9 when μ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

HWsolutions4 - HOMEWORK#4 SOLUTIONS IE121 8-7 = P X > 185...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online