Statistics - Speech or Presentation Example

Mathematics Topic: Statistics Presented by: On 26th may, Solution to Question One n=100 mean=30 sd=8 CI=95% Zα/2=1.96I. Between 28.4 and 31.6standard error= 8/10=0.8 ᵶ1=(28.4-30)/0.8= -2 Therefore area = -0.4772ᵶ2= (31.6-30)/0.8= 2 Therefore area = 0.4772P( 28.4≤µ≤31.6)= P( 31.6) – P( 28.4)= 0.4772- ( -0.4772)= 0.9544 0.9544*100=95.44%, thus it is 95% sure that the population mean lies between 28.4 and 31.6II. Between 27.0 and 33.0ᵶ1=(27.0-30)/0.8= -3.75 therefore area = -0.

4999 ᵶ2 = (33.0-30)/0.8 = 0.375 therefore area = 0.4999 P(27.0 and 33.0)= 0.4999-(-0.4999)=0.99980.9998*100%= 99.98%, thus it is 99.98% confident that the population mean lies between 27.0 and 33.0III. Between 26.1 and 34.2ᵶ1=(26.1-30)/0.8=-4.875 therefore the area = -0.5 ᵶ2 = (34.2-30)/0.8= 0.5P( 26.1 and 34.2) = 0.5 -(-0.5)=11*100=100%, thus it is 100% confident that population mean lies between 26.1 and 34.2IV. Between 24.1 and 35.9ᵶ1=(24.1-30)/0.8= -7.375 therefore area = - 0.

5 ᵶ2=(35.9-30)/0.8=7.375 therefore area = 0.5P(24.1≤µ≤35.9)=0.5-(-0.5)=11*100%=100%, thus it is 100% confident that population mean lies between 24.1 and 35.9Solution to Question TwoMean=12.5 n=40 sd=4 ᵶα/2=2.575 standard error== 0.6325I. Between 11.5 and 13.5ᵶ1=(11.5-12.5)/0.6325= -1.5810 therefore the area is -0.4429ᵶ2=(13.5-12.5)/0.6325= 1.5810 therefore the area is 0.44290.4429-(-0.4429) =0.88580.8858*100%=88.58% thus 11.5 and 13.5 does not lie within 99% interval.II. Between 10.9 and 14.1ᵶ1= (10.9-12.5)/0.6325= - 2.5296 therefore the area is -0.

4943ᵶ2=(14.1-12.5)/0.6325= 2.5296 therefore the area is 0.49430.4943-(-0.4943)= 0.98860.9886*100%=98.86% thus 10.9 and 14.1 does not lie within the 99% confidence interval.III. Between 10.1 and 15.1ᵶ1= (10.1-12.5)/0.6325= -3.794 therefore the area is -0.4999ᵶ2= (15.1-12.5)/0.6325= 4.1106 therefore the area is 0.50.5-(-0.4999) = 0.99990.9999*100%=99.99% therefore it is confident that 10.1 and 12.5 lies within 99% confidence interval.IV. Between 7.4 and 14.2ᵶ1=(7.4-12.5)/0.6325= -8.0632 therefore area is -0.5ᵶ2=(14.2-12.5)/0.6325= 2.

6877 therefore area is 0.49640.4964-(-0.5)= 0.99640.9964*100%=99.64% therefore 7.4 and 14.2 lies within 99% confidence interval.Solution to Question Threen=16 d.f=15 ᵶt from the tableα=0.10 and the confidence interval is 90%.From the table; d.f=15, α=0.10 therefore ᵶt=1.341Solution to Question Fourmean=150 n=36 d.f=n-1=35 s=40 c.i=80%standard error ==6.67I. Between 135.6 and 148.4ᵶ1=(135.6-150)/6.67= -2.1589 therefore area is -0.4846ᵶ2=(148.4-150)/6.67= -0.2398 therefore area is -0.0948-0.0948-(-0.4846)=0.38980.

3898*100%= 38.98% therefore 135.6 and 148.4 does not lie within 80% confidence interval.II. Between 154.2 and 175.8ᵶ1=(154.2-150)/6.67= 0.6296 therefore the area is 0.2357ᵶ2=(175.8-150)/6.67= 3.868 therefore the area is 0.49990.4999-0.2357=0.26420.2642*100%=26.42% therefore 154.2 and 175.8 does not lie within 80% confidence intervalIII. Between 141.3 and 158.7ᵶ1=(141.3-150)/6.67= -1.304 therefore area is -0.4032ᵶ2=(158.7-150)/6.67= 1.304 therefore area is 0.40320.4032-(-0.4032)=0.80640.

8064*100%=80.64% therefore 141.3 and 158.7 lies within 80% confidence intervalIV. Between 162.3 and 187.5ᵶ1=(162.3-150)/6.67= 1.8440 therefore area is 0.4671ᵶ2=(187.5-150)/6.67= 5.622 therefore area is 0.50.5-0.4671=0.03290.0329*100%=3.29% therefore 162.3 and 187.5 does not lie within 80% confidence intervalSolution to Question FiveI. Between 0.26 to 0.64ᵶ1== 4.923 therefore the area= -0.5ᵶ2== 3.58060.4998-(-0.58) =99.98% , Its 99.98%confident that 0.26 and 0.64 lies within 90% confident interval.II. 0.39 to 0.57ᵶ1= =-2.

0141, therefore the area is = -.04778ᵶ2==2.0141, the area is=0.4778=0.4778-(-0.4778)=0.95560.9556*100= 95.56%, its 95.56% that 0.39 to 0.57 lies within 90% confident interval.III. 0.41 to 0.52ᵶ1==-1.56652, therefore the area is =-0.4418ᵶ2==0.89515, therefore the area is =0.31330.3133-(-0.4418)=0.7551*100=75.51% , its 75.51% confident that 0.41 and 0.52 lies within 90%confident interval.IV. 0.28 to 0.71ᵶ1==-4.475 therefore area is -0.5ᵶ2==5.14714 therefore area is 0.50.5-(-0.5)=1 therefore it is 100% confident that 0.28 and 0.71 lies within 90% confidence interval.

Solution to Question SixI. An example of estimation of population meanA producer wishes to determine the mean daily output level for his products. He selects a sample of 100 days and determines the average output is 112 tones. Past experience has shown that the population standard deviation is 50 tones. Determine the interval in which the mean lies at 95% level of confidence interval.sd=50 n=100 mean=12µ=112±1.96 ()102.2≤µ≤128.8 Therefore the producer is 95% confident that the mean daily output lies between 102.2 and 121.8II.

An example of estimation of population proportionOut of the patients data a certain doctor treated, 45% never died. If a sample of 80 patients is selected from this population of patients, what is the probability that more than 40 patients never died?π=0.45 1-π=0.55 n=80 P==0.5ᵶ==0.89 therefore area is 0.3133P(x>0.5)=0.5-0.3133=0.18670.1867*100%=18.67%Works CitedKendall, M. G., Stuart, A., Ord, J. K., Arnold, S. F., & Hagan, A. (1994-2004). Kendalls advanced theory of statistics (6th ed.). London: Edward Arnold.

Kincaid, C. D., Wackerly, D. D., Mendenhall, W., & Scheaffer, R. L. (1996). Student solutions manual for Wackerly, Mendenhall, and Scheaffers Mathematical statistics with applications. Belmont: Duxbury Press.