Statistical Analysis and Hypothesis Testing in Various Research Scenarios

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Statistics 51.A researcher is comparing the reaction times of 10 year olds who playvideogames and 10-year-olds who do not. Random samples from eachgroup provide the results shown below.Reaction times of 10-year-oldsPlayers ofVideo GamesNon-PlayersOf Video GamesSample size1510Mean reaction time(seconds)0.500.52Standard deviationof reaction times(seconds)0.010.02Assume that the reaction times for both populations are normally distributedwith the same population standard deviation.a.Construct a 98% confidence interval for the difference between thepopulation means of the reaction times.Note: Express your answer to 5 decimal places of accuracy.𝑠𝑝=ට(𝑛1−1)𝑠12+(𝑛2−1)𝑠22𝑛1+𝑛2−2=ට(15−1)(0.01)2+(10−1)(0.02)215+10−2=0.01474.The98% confidence interval for the difference between thepopulation means of the reaction times

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቎(𝒙ഥ𝟏−𝒙ഥ𝟐)±𝒕𝜶𝟐⁄,𝒏𝟏+𝒏𝟐−𝟐𝒔𝒑ඨ𝟏𝒏𝟏+𝟏𝒏𝟐቏=቎(𝟎.𝟓𝟎−𝟎.𝟓𝟐)±𝒕𝟎.𝟎𝟏,𝟐𝟑(𝟎.𝟎𝟏𝟒𝟕𝟒)ඨ𝟏𝟏𝟓+𝟏𝟏𝟎቏=[−𝟎.𝟎𝟐𝟎𝟎𝟎±(𝟐.𝟒𝟗𝟗𝟖𝟕)(𝟎.𝟎𝟎𝟔𝟎𝟐)]=[−𝟎.𝟎𝟐𝟎𝟎𝟎±𝟎.𝟎𝟏𝟓𝟎𝟒]=[−𝟎.𝟎𝟑𝟓𝟎𝟒,−𝟎.𝟎𝟎𝟒𝟗𝟔]b.Using a 2.5% significance level, can the researcher conclude that the meanreaction time of players of video games is less than the meanreaction timeof non-players of video games? Formulate and test the appropriatehypotheses. Use the critical value approach.Null hypothesis,H0:μ1≥μ2Alternative hypothesis,Ha:μ1<μ2Level of significance,α= 0.025The critical value for this left-tailed t-test with𝒏𝟏+𝒏𝟐−𝟐=𝟐𝟑degrees of freedom at 2.5% significance level is–2.0687.Test statistic,𝒕=(𝒙ഥ𝟏−𝒙ഥ𝟐)𝒔𝒑ට𝟏𝒏𝟏+𝟏𝒏𝟐=𝟎.𝟓𝟎−𝟎.𝟓𝟐(𝟎.𝟎𝟏𝟒𝟕𝟒)ට𝟏𝟏𝟓+𝟏𝟏𝟎=−𝟎.𝟎𝟐𝟎.𝟎𝟎𝟔𝟎𝟐=−𝟑.𝟑𝟐𝟑𝟔Since–3.3236 <–2.0687, reject the null hypothesis.Therefore, at 2.5% significance level, the researcher canconclude thatthe mean reaction time of players of videogames is less than the meanreaction time of non-playersof video games.2.Is there any difference between the proportion of Americans who wish theywere rich and the proportion of Canadians who wish they were rich? A randomsample of 1,000 Americans showed that550 wish they were rich, while arandom sample of 500 Canadians showed that 225 wished they were rich.a.Find a 97.5% confidence interval for the difference between the population

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proportions who wish they were rich.Note: Express your answer to 4 decimal places of accuracy.𝑝̂1=𝑋1𝑛1=5501,000=0.55𝑝̂2=𝑋2𝑛2=225500=0.45The 97.5% confidence interval for the difference between thePopulation proportions who wish they were rich:቎(𝒑ෝ𝟏−𝒑ෝ𝟐)±𝒁𝜶𝟐⁄ඨ𝒑ෝ𝟏(𝟏−𝒑ෝ𝟏)𝒏𝟏+𝒑ෝ𝟐(𝟏−𝒑ෝ𝟐)𝒏𝟐቏=቎(𝟎.𝟓𝟓−𝟎.𝟒𝟓)±𝒁𝟎.𝟎𝟏𝟐𝟓ඨ𝟎.𝟓𝟓(𝟏−𝟎.𝟓𝟓)𝟏,𝟎𝟎𝟎+𝟎.𝟒𝟓(𝟏−𝟎.𝟒𝟓)𝟓𝟎𝟎቏=[𝟎.𝟏𝟎𝟎𝟎±(𝟐.𝟐𝟒)(𝟎.𝟎𝟐𝟕𝟐)]=[𝟎.𝟏𝟎𝟎𝟎±𝟎.𝟎𝟔𝟏𝟎]=[𝟎.𝟎𝟑𝟗𝟎,𝟎.𝟏𝟔𝟏𝟎]b.Test at the 1% level of significance, whether the twopopulation proportionsare significantly different. Formulate and test the appropriate hypotheses.Use the critical value approach.Null hypothesis,H0:p1=p2Alternative hypothesis,Ha:p1≠p2Level of significance,α= 0.01The critical valuesfor thistwo-tailedz-test at1% significance levelare–2.576 and 2.576.Test statistic,𝒛=൫𝑝ෝ1−𝑝ෝ2൯ඨ𝒑ෝ𝟏൫𝟏−𝒑ෝ𝟏൯𝒏𝟏+𝒑ෝ𝟐൫𝟏−𝒑ෝ𝟐൯𝒏𝟐=𝟎.𝟓𝟓−𝟎.𝟒𝟓ට𝟎.𝟓𝟓(𝟏−𝟎.𝟓𝟓)𝟏,𝟎𝟎𝟎+𝟎.𝟒𝟓(𝟏−𝟎.𝟒𝟓)𝟓𝟎𝟎=𝟎.𝟏𝟎𝟎.𝟎𝟐𝟕𝟐=𝟑.𝟔𝟖

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