8.48 Research Exercise

RES 342 Ch. 8.48 Xbar= 346.5 X= 170.38 A.) 95% authorisation interval for accredited mean Degrees of freedom: n-1= 20-1= 19 t= 2.093 Lower outset= xbar t(s/?n) 346.5-2.093 (170.38/ ?20) =266.76 Upper limit= xbar + t(s/?n) =346.5+ 2.093(170.38 /?20) =462.24 B.) Is normality an issue here? maybe only because the domain falls between both the belt along and lower limit- the upper limit is the entire page term the lower is no advertisement C.) standard size undeniable to obtain an error of ±10 squ be millimeters with 99 percent confidence? n= (zs/E)^2 n= (2.5758*170.38/10)^2 n=1926.02 D.) Is this a reasonable requirement? No, this is a trouble because the sustain itself only has 1591 pages thus a lower confidence train such as 95% would be to a greater extent appropriate. Ch. 8.64 A.) Construct 90% confidence interval for counterweight of unpopped kernels Sample proportion: P= x/n=0.1113 Z cling to corresponding to a=0. 10 is 1.645 =(0.112-1.645 * Sqrt[(0.1113*0.8887)/773] And = (0.1113+1.645*Sqrt [0.1113*0.8887/773]) Therefore (0.0927, 0.1299) B.) Normality presumptuousness Check n*p=773*0.1113=86 n(1-p)=777*.8887=687 86 & 687 are greater than 5 therefore the normality assumption is cheery C.) Very Quick Rule This does non work because the p value is equal to 0.1113 which is not underweight to 0.5 which the very affectionate rules requires p to be. D.) Is sample typical? No this sample is not typical because the sample is not randomIf you wish to get a full essay, order it on our website: OrderCustomPaper.com

If you want to get a full essay, visit our page: write my paper