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Statistical Hypothesis Testing and Confidence Intervals: A Comprehensive Analysis - Document preview page 1

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Statistical Hypothesis Testing and Confidence Intervals: A Comprehensive Analysis

Analyzes hypothesis testing and confidence intervals.

Olivia Smith
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Statistical Hypothesis Testing and Confidence Intervals: A Comprehensive Analysis - Page 1 preview image1Statistical Hypothesis Testing and Confidence Intervals: A ComprehensiveAnalysisMake sure your answers are as complete as possible and show your work/argument.When there are calculations involved, you should show how you come up with youranswers with critical work and/or necessary tables.You must show why you choosecertain answer for true-or-false and multiple choice questions.Answers that comestraight from program software packages will not be accepted.1. (2 pts)Trueor False: In a right-tailed test, the test statistic is 1.5. If we know P(X < 1.5) =0.96, then we reject the null hypothesis at 0.05 level of significance. (Justify for full credit)BecauseP(X2. (2 pts) True orFalse:If a 99% confidence interval contains 1,thenthe95% confidence intervalfor the same parametermustcontain1. (Justify for full credit)Because 99% confidence is wider than 95% confidence interval so 99%confidence intervalcontains 1 but 95% confidence interval may not contain 1for the same parameter.3. (2 pts)Which of the following could reduce the rate of Type I error?(Justify for full credit)a. Making the significant level from 0.01 to 0.05b. Making the significant level from 0.05 to 0.01c. Increase the β leveld. Increase the powerBecause level of significance is actually the type I error in testing of hypothesis so we need toreduce the level of significance from 0.05 to 0.01 to reduce the type I error.4. (2 pts)Threehundred students took a chemistry test. You sampled50students to estimate theaverage score and the standard deviation. How many degrees of freedom were there in theestimation of the standard deviation?(Justify for full credit)a.50b.49c.300d.299Here the sample size n=50 so the degree of freedom is =50-1=49.(For Questions5&6)Mimi was the 5th seed in 2015UMUC Tennis Open that took place inAugust. In this tournament, she won75of her 100 serving games.
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Statistical Hypothesis Testing and Confidence Intervals: A Comprehensive Analysis - Page 3 preview image25.(2pts)Find a 90% confidence interval estimate of the proportion ofserving games Mimi won.(Show workand round the answer to three decimal places)Here sample proportion p=.90% confidence interval for the proportion is(p-Z(1-0.9)/2*p+Z(1-0.9)/2*)=( p-Z0.05*, p+Z0.05*)=(0.75-1.645*0.75+1.645*)=(0.679, 0.821)6.(5pts)According toUMUC Sports Network,Mimiwins80% of the serving games in her 5-year tennis career.In order to determine ifthis tournamentresultisworsethanher career record of80%.Wewould like to perform the following hypothesis test:(a) (2 pts) Find the test statistic. (Show work and round the answer to two decimal places)Sample size =n=100 and sample proportion P=0.75.Under the null hypothesis the test statistics isZ=(b)(2pts) Determine theP-value for this test. (Show workand round the answer to threedecimal places)The P value is P(Z<-1.25)=0.106
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