In an experiment the sorption was measured for a variety of hazardous organic chloroalkane solvents. The sample measurements were; 1.18, 1.35, 0.57, 1.16, 1.12, 0.91, 0.83, 0.43 Assuming that measurements are normally distributed. i) What is the mean of the sample for sorption of chloroalkane solvents. ii) What is the standard error of the mean for the sample of organic chloroalkane solvents? iii) What the margin of error of the 99% confidence intervals for the population mean of sorption of chloroalkane solvents. iv) What is the lower limit for the 99% confidence intervals for the population mean of sorption of chloroalkane solvents.

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In an experiment the sorption was measured for a variety of hazardous organic chloroalkane solvents. The sample measurements were; 1.18, 1.35, 0.57, 1.16, 1.12, 0.91, 0.83, 0.43 Assuming that measurements are normally distributed.  i) What is the mean of the sample for sorption of chloroalkane solvents.    ii) What is the standard error of the mean for the sample of organic chloroalkane solvents?    iii) What the margin of error of the 99% confidence intervals for the population mean of sorption of chloroalkane solvents.    iv) What is the lower limit for the 99% confidence intervals for the population mean of sorption of chloroalkane solvents.

Pansdorfp6 · Accepted Answer

Step 1  Solution:  x(x−x¯)(x−x¯)21.180.23620.055791.350.40620.1649980.57−0.37380.1397261.160.21620.0467421.120.17620.0310460.91−0.03380.0011420.83−0.11380.012950.43−0.51380.26399  n=8 Sample size  ∑x=7.55  Step 2  i) To find the mean of the sample for sorption of chloroalkane solvents.  x―=∑xn  x―=7.558=0.9438  The mean of the sample for sorption of chloroalkane solvents is 0.9438   ii)  To find the  sample standard deviation for sorption of chloroalkane solvents.  ∑(x−x―)2=0.7163388  s=∑(x−x―)2n−1  s=0.7163887  s=0.1023411  s=0.3199  The standard error of the mean for the sample of organic chloroalkane solvents is   sx―=sn  sx―=0.31998  sx―=0.1131  The standard error of the mean for the sample of organic chloroalkane solvents is   0.1131  iii)  The margin of error at 99% confidence intervals for the population mean of sorption of chloroalkane solvents is  Margin of error=E=tα2, n−1×sx―  At aplha=0.01  tα2, n−1=3.499 From t table  Margin of error =E=3.499×1131  Margin of error =E=0.3957369  Margin of error =E=0.3957  The margin of error at 99% confidence intervals for the population mean of sorption of chloroalkane solvents is 0.3957  iv) The lower limit for the 99% confidence intervals for the population mean of sorption of chloroalkane solvents is  Lower Limit =x―−E  Lower Limit =0.9438−0.3957  Lower Limit =0.5481  The lower limit for the 99% confidence intervals for the population mean of sorption of chloroalkane solvents is 0.5481

sirpsta3u · Accepted Answer

Step 1 95% confidence interval when sumple mean and variance known P(x―−tα2(3n)&amp;lt;μ&amp;lt;x―+tα2(3n)) x―=∑xin x: measurments =1.658 x―=0.9562 =∑(xi−x―)2n−1 =0.3321 tα2=tn−1, α=0.1 tα2=1.845 Step 2 Morgin of error =tα2(3n) =1.895(0.33218) Mergin of error =0.2250 Step 3 Upper limit of 90% is x―+morgin of error x―+tα2(3n) 0.9562+0.2250 ∵1.787015 Upper limit for 90% is 1.787015

In an experiment the sorption was measured for a variety of

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