Homeowrk 7 Solution

# Homeowrk 7 Solution - HW7 1. A study was conducted of 90...

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Unformatted text preview: HW7 1. A study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment period. The previous treatment regime had produced an average increase of μ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of μ in comparison to the previous treatment. The data yielded x =2.17 and s=1.05. Using α =.05, what conclusion can you draw about the research hypothesis? (Conduct the hypothesis test.) 90; 2; 2.17; 1.05; .05 1) : 2 : 2 2) ( 30, , ) 2.17 2 1.535 / 1.05 / 90 3) ( .05) a n x s Hypotheses H versus H Test Statistic n no assumption of normality of population unknown Case II x z s n Rejection Region Z z Z z α α μ α μ μ σ μ α = = = = = = ≥ =-- ∴ = = = = ≥ = ≥ .05 1.645 4) 1.535 (1.535 1.645), .05. z Conclusion The computed value z does not fall in the rejection region so H cannot be rejected at significance level The data does not give strong support to the claim that the new treatment had i = = = < ∴ . ncreased the value of in comparison to the previous treatment μ 2. Set up the rejection region based on t for the following conditions with α =.05: (a) : : , 15 a H versus H n μ μ μ μ = < = , 1 .05,14 15; .05 : 1.761 1.761 n n Rejection Region T t t T α α- = = ≤ - = - = - ∴ ≤ - (b) : : , 23 a H versus H n μ μ μ μ = ≠ = /2, 1 /2, 1 .025,22 .025,22 23; .05 : 2.074 2.074 n n n Rejection Region T t or T t T t or T t T or T α α α-- = = ≤ - ≥ ≤ - ≥ ∴ ≤ - ≥ (c) : : , 6 a H versus H n μ μ μ μ = = , 1 .05,5 6; .05 : 2.015 2.015 n n Rejection Region...
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## This note was uploaded on 03/28/2011 for the course AMS 110 taught by Professor N/a during the Spring '08 term at SUNY Stony Brook.

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Homeowrk 7 Solution - HW7 1. A study was conducted of 90...

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