STATS
Stats_Material47

# Stats_Material47 - Q1 Based on a sample of 50 x-values...

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Q1 Based on a sample of 50 x -values having mean 35.36 and standard deviation 4.26, (a) test at the 0.05 level of significance the null hypothesis :  = 34 versus the alternative :   34. (b) find a 95% confidence interval for the population mean. (c) determine if the results to part (a) and (b) are mutually consistent. (d) use Minitab to give the p -value for the test in (a).  Ans. a.   Set Up Hypothesis Null Hypothesis H0: U=34 Alternate Hypothesis H1: U!=34 Test Statistic Population Mean(U)=34 Given That X(Mean)=35.36 Standard Deviation(S.D)=4.26 Number (n)=50 we use Test Statistic (Z) = x-U/(s.d/Sqrt(n)) Zo=35.36-34/(4.26/Sqrt(50) Zo =2.2574 | Zo | =2.2574 Critical Value The Value of |Z α| at LOS 0.05% is 1.96 We got |Zo| =2.2574 & | Z α | =1.96 Make Decision Hence Value of | Zo | > | Z α| and Here we Reject Ho Sample mean is not equal to the population mean b.   Confidence Interval CI = x ± Z a/2 * (sd/ Sqrt(n)) Where, x = Mean sd = Standard Deviation a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=35.36 Standard deviation( sd )=4.26 Sample Size(n)=50 Confidence Interval = [ 35.36 ± Z a/2 ( 4.26/ Sqrt ( 50) ) ] = [ 35.36 - 1.96 * (0.602) , 35.36 + 1.96 * (0.602) ] = [ 34.179,36.541 ] Sample mean is not equal to the population mean c.   Result in both a and b are mutually consistent, since both results in different in population mean d. P-Value : Two Tailed ( double the one tail ) - Ha : ( P != 2.2574 ) = 0.024 Q2 The data set CHS\NBA.MTP contains a number of variables for guards in the National Basketball Association. ( you can find data set here: ) (a) For the variable PPG, meaning points per game, test the null hypothesis PPG = 11 against the alternative PPG  11 using the 0.05 level of significance. The data file has n = 105, = 10.70, s = 6.058. (b) Find a 95% confidence interval for  PPG and reconcile your conclusion in (a) with this interval. Ans. a.   Test Used: Z-Test For Single Mean Set Up Hypothesis Null Hypothesis H0: U=11 Alternate Hypothesis H1: U!=11 Test Statistic Population Mean(U)=11 Given That X(Mean)=10.7 Standard Deviation(S.D)=6.058 Number (n)=105 we use Test Statistic (Z) = x-U/(s.d/Sqrt(n)) Zo=10.7-11/(6.058/Sqrt(105)

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Zo =-0.5074 | Zo | =0.5074 Critical Value The Value of |Z α| at LOS 0.05% is 1.96 We got |Zo| =0.5074 & | Z α | =1.96 Make Decision Hence Value of |Zo | < | Z α | and Here we Do not Reject Ho b. Confidence Interval CI = x ± Z a/2 * (sd/ Sqrt(n)) Where, x = Mean sd = Standard Deviation a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=10.7 Standard deviation( sd )=6.058 Sample Size(n)=105 Confidence Interval = [ 10.7 ± Z a/2 ( 6.058/ Sqrt ( 105) ) ] = [ 10.7 - 1.96 * (0.591) , 10.7 + 1.96 * (0.591) ] = [ 9.541,11.859 ] Q3 Suppose that a sample of 200 accounts receivable entries at a large mail-order business had a mean price of \$846.20 and a standard deviation of \$1,840.80. Give a 95% confidence interval for the population mean. Be sure to state any assumptions that you use.
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