STAT 200 Quiz 3 Section 6380

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1 STAT 200 QUIZ 3 Section 638 0 Summer 201 5 ( For Questions 1 & 2 ) Mimi was the 5th seed in 201 4 UMUC Tennis Open that took place in August. In this tournament, she won 80 of her 100 serving games. Based on UMUC Sports Network, she wins 75 % of the serving games in her 5 - year tennis career. 1. ( 2 pts) Find a 9 0 % confidence interval estimate of the proportion of serving games Mimi won . (Show work and round the answer to three decimal places ) 9 0 % 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. ( 5 pts) In order to determine if this tournament result is better than her career record of 75 % . We would like to perform the following hypothesis test: 𝐻 0 : 𝑝 = 0 . 75 𝐻 ௔ : 𝑝 > 0 . 75 ( a ) (2 pts) Find the test statistic. (Show work and round the answer to two decimal places) z= (.8 - .75)/ sqrt(.75*.25/100) = .05/.0433 = 1.15 ( b ) ( 2 pts) Determine the P - value for this test. (Show work and round the answer to three decimal places) P(z> 1.15)= 1 - P(z<1.15)= 1 - .875 = .125 ( c ) ( 1 pt) Is there sufficient evidence to justify the rejection of 0 H at the α = 0 . 02 level? Explain. T here is in sufficient evidence to justify the rejection of 0 H at the α = 0 . 02 level as the p - value of .125> α =.02. 3 . ( 5 points) The SAT scores are normally distributed. A simple random sample of 2 25 SAT scores has a sample mean of 15 0 0 and a sample standard deviation of 3 00 . (a) (1 pt) What distribution will you use to determine the critical value? Why? z ~ N(0,1) SAT scores are normally distributed and n=225 ( b ) ( 3 pts) Construct a 95% confidence interval estimate of the mean SAT score. (Show work and round the answer to two decimal places ) 95% confidence interval for 𝜇 : ( xbar – 1.96*300/sqrt(225) , xbar + 1.96*300/sqrt(225))=(1500 -
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