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STAT 200 Quiz 3 Section 6380

A quiz covering key statistical concepts for STAT 200 students.

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1STAT 200 QUIZ3Section6380Summer2015(For Questions1&2)Mimi was the 5th seed in 2014UMUC Tennis Open that took place inAugust. In this tournament, she won80of her 100 serving games. Based on UMUC SportsNetwork, she wins75% of the serving games in her 5-year tennis career.1.(2pts)Find a 90% confidence interval estimate of the proportion ofserving games Mimiwon.(Show workand round the answer to three decimal places)90% confidence interval estimate of the proportion of serving games Mimi won= (.8-1.645*sqrt(.75*.25/100 ,.8+ 1.645*sqrt(.75*.25/100 )= (.729, .871)2.(5pts)In order to determine ifthis tournamentresultisbetter thanher career record of75%.Wewould like to perform the following hypothesis test:𝐻0:𝑝=0.75𝐻௔:𝑝>0.75(a) (2 pts) Find the test statistic. (Show work and round theanswer to two decimal places)z= (.8-.75)/sqrt(.75*.25/100) = .05/.0433 =1.15(b)(2pts) Determine theP-value for this test. (Show work and round the answer to threedecimal places)P(z> 1.15)= 1-P(z<1.15)= 1-.875 =.125(c)(1pt) Is there sufficient evidence to justify the rejection of0Hat theα=0.02level?Explain.Thereis insufficient evidence to justify the rejection of0Hat theα=0.02level as thep-valueof.125>α=.02.3.(5points)The SAT scores are normally distributed.A simple random sample of225SATscores has asamplemean of 1500andasamplestandard deviation of 300.(a) (1 pt) What distribution will you use to determine the critical value? Why?z~N(0,1)SAT scores are normally distributedand n=225(b)(3pts)Construct a 95% confidence interval estimate of themean SAT score.(Show workand round the answer to two decimal places)95% confidence intervalfor𝜇:(xbar–1.96*300/sqrt(225) , xbar + 1.96*300/sqrt(225))=(1500-

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Document Details

University
University of Maryland
Course
STAT 200
Subject
Statistics

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