hw.prop.CI.Hypo.pVal.STS.KEY

# 55045 0099 n 25 pcv 055 2330099 055 023

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Unformatted text preview: p < 0.55 2. Determine critical value ^ ^ pcv - p pcv - 0.55 KEY: Zcv = -2.33 = p(1- p) 0.55(0.45) 0.099 = n 25 ^ pcv = 0.55 – 2.33(0.099) = 0.55 – 0.23 = 0.32 3. 4. 5. 8. 8a. 8b. State decision rule KEY: Reject Ho if ^sample < ^cv = 0.32 p p p State decision KEY: Since ^sample = 9/25=0.36 NOT < ^cv = 0.32 can NOT reject Ho p State conclusion KEY: the population proportion p is probably ≥ 0.55 We'll test at α = 0.05 whether the population proportion of airline passengers checking baggage is at most 0.55 . A sample of 25 passengers showed 11 checked their baggage. Determine whether it is OK to use the Normal distribution for this problem. KEY: OK to use Z if np>5 and n(1-p)>5: 25(0.55)=13.75 > 5, 25(0.45)=11.25 > 5 so OK to use Z Hypothesis testing using the STS method: Show all work including the raw formula with symbols (no numbers). No credit unless proper subscripts are used everywhere. Use 3 decimal places for the standard deviation. Draw a graph and fill it out completely. 1. State Hypotheses KEY: Ho p ≤ 0.55 H1: p > 0.55 2. Determine critical value KEY: Zcv = 1.645 3. State decision rule: KEY: reject Ho if Zsample > Zcv = 1.645 4. State decision KEY: Since Zsample = ^ psample - p KEY: Zsample = Zsample = p(1- p) n [11/25=0.44] - 0.55 -0.11 = 0.099 = -1.11 0.55(0.45) 0.099 = 25 Since Zsample = -1.11 NOT > Zcv = 1.645, can NOT reject Ho 5. State conclusion KEY: the population proportion p probably IS ≤ 0.55 8c. Note that the p-value is > 0.5000: area for -1.11 is 0.3665, so p-value = 0.5000 + 0.3665 = 0.8665 . p-value is on the same side as alpha – it has the same "tail" as alpha....
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## This note was uploaded on 01/27/2013 for the course STAT 385 taught by Professor Szatrowski during the Spring '08 term at Rutgers.

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