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Statistics Assignment - T Test - Desklib

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Added on  2020-03-23

Statistics Assignment - T Test - Desklib

   Added on 2020-03-23

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Surname 1StatisticsNameThe Name of the Class (Course)Professor (Tutor)The Name of the School (University)The City and State where it is locatedDate
Statistics Assignment - T Test - Desklib_1
Surname 2QUESTION 1. T-test [20 marks]a)The most appropriate test is two sample tests, since we are testing the difference between twoindependent averages. In this case, the paired t-test cannot be applied since the weight of theplayers is not taken from the same sample at two different occasions or time. The sample isalso less than 30, and the population standard deviation is unknown. Therefore, two sample t-test is the most appropriate test to be applied.The only reason thatcan make the test inappropriate is when the data are very skewed. Also, when the sample issufficiently large (n > 30), the z-statistics will be used. Lastly, when the sample does notcome from a normally distributed population. b)H0: the average weight of a footballer and basketball player is equal, Versus Ha: the averageweight of football players is at least 45 pounds more than the average weight of basketballplayers. Two-Sample T-Test and CI: weight, sport Two-sample T for weightSport N Mean StDev SE MeanBasketball 19 205.8 12.9 3.0Football 19 258.8 12.4 2.8Difference = μ (basketball) - μ (football)Estimate for difference: -53.0095% upper bound for difference: -46.07T-Test of difference = 45 (vs <): T-Value = -23.90 P-Value = 0.000 DF = 35The results show that there is enough evidence to reject the null hypothesis (t (35) = -23.90, p< .000) [ CITATION Moo13 \l 1033 ]. This means that the claim that the average weight of
Statistics Assignment - T Test - Desklib_2
Surname 3football players is at least 45 pounds more than the average weight of basketball players istrue at the 95% level of confidence. c)The 95% upper bound for difference is -46.07, so the C.I is (-46.07, -59.93). This shows thatwe are 95% confident that the average population difference of the basket players and thefootball players is between the upper and lower bound of the interval.d)H0: Football data are normally distributed.Ha: the football data are not normally distributed. 290280270260250240230999590807060504030201051Mean258.8StDev12.38N19RJ0.982P-Value>0.100footballPercentProbabilityPlotoffootballNormalThe results show that RJ = 0.982, p-value > 0.100, which suggests that the null hypothesisshould not be rejected [ CITATION YuQ16 \l 1033 ]. In particular, there is enough evidenceto shows that we cannot conclude that the football data do not follow a normally distribution.e)The data are as follows.footbaltransforme
Statistics Assignment - T Test - Desklib_3
Surname 4ld football weight245-1.114172620.259407255-0.30619251-0.62938244-1.194972761.390592240-1.518172650.501804257-0.14459252-0.548582821.875385256-0.22539250-0.710182640.4210052700.9057982751.309793245-1.114172751.309793253-0.46778The plot is as follows.210-1-2543210Mean4.616190E-16StDev1N19transformedfootballweightFrequencyHistogramoftransformedfootballweightNormalThe plot supports that the data follows a student’s t-distribution since the average isapproximately zero, and the standard deviation is equal to one.
Statistics Assignment - T Test - Desklib_4

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