In an experiment the sorption was measured for a variety of

pierdoodsu

pierdoodsu

Answered question

2021-12-17

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.

Answer & Explanation

Pansdorfp6

Pansdorfp6

Beginner2021-12-18Added 27 answers

Step 1
Solution:
x(xx¯)(xx¯)21.180.23620.055791.350.40620.1649980.570.37380.1397261.160.21620.0467421.120.17620.0310460.910.03380.0011420.830.11380.012950.430.51380.26399
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.
(xx)2=0.7163388
s=(xx)2n1
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, n1×sx
At aplha=0.01
tα2, n1=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 =xE
Lower Limit =0.94380.3957
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

sirpsta3u

Beginner2021-12-19Added 42 answers

Step 1
95% confidence interval when sumple mean and variance known
P(xtα2(3n)<μ<x+tα2(3n))
x=xin
x: measurments
=1.658
x=0.9562
=(xix)2n1
=0.3321
tα2=tn1, 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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?