I know how to perform the tests that its asking us to use in the problem, but pe

Question

I know how to perform the tests that its asking us to use in the problem, but performing the math with the concentrations in ppm is throwing me off, especially since part c) represents the mean and standard deviation as percentages. Please help!

gem. Find the true mass of the aliquot of pentane. Repeated measurements of the lead concentration in a contaminated soil sample gave values of 856.4, 890.7, 862.3, 844.2, and 852.5 ppm. (a) What are the mean concentration, standard deviation, and 95% confidence interval in the mean Pb concentration? (b) Use the Q-test to determine if the highest value can be discarded. (c) A second soil sample analyzed 6 replicate times for Pb gave a mean of 0.0882% and standard deviation of0.730%. At the 95% confidence level, do the two soil samples differ?

Explanation / Answer

Part a

For the given data, we have

Sample size = n =5

Sample mean = Xbar = 861.22

Sample standard deviation = S = 17.74252

Now, we have to find the 95% confidence interval for the population mean.

Confidence interval = Xbar ± t*S/sqrt(n)

Degrees of freedom = n – 1 = 5 – 1 = 4

Confidence level = 95%

Critical t value = 2.7764 (by using t-table/excel)

Confidence interval =861.22 ± 2.7764*17.74252/sqrt(5)

Confidence interval =861.22 ± 2.7764* 7.934695961

Confidence interval =861.22 ± 22.0302

Lower limit = 861.22 - 22.0302 = 839.19

Upper limit = 861.22 + 22.0302 = 883.25

Confidence interval = (839.19, 883.25)

Part b

Here, we have to perform Dixons Q-test for testing whether outlier can be discarded or not.

Null hypothesis: H0: Highest value cannot be discarded.

Alternative hypothesis: Ha: Highest value can be discarded.

We assume level of significance = = 0.05

Test statistic is given as below:

Q = (x2 – x1)/(xn – x1)

Where, x1 = smallest value

x2 = second smallest value

xn = largest value

From given data, we have

x1 = smallest value = 844.2

x2 = second smallest value = 852.5

xn = largest value = 890.7

Q = (852.5 - 844.2) / (890.7 - 844.2)

Q = 0.178495

We are given n = 5, = 0.05, so from Dixon’s Q table, we have

Critical Q value = 0.710

Statistic Q = 0.178495 < Critical Q = 0.710

So, we do not reject the null hypothesis that highest value cannot be discarded.

So, we conclude that the highest value cannot be discarded.

Part c

For the first sample, we have

Sample size = n =5

Sample mean = Xbar = 861.22 ppm = 861.22 / 10000 = 0.086122%

(Divide value by 10000 for conversion of ppm into percentage.)

(ppm = part per million)

(1 million = 1,000,000, if we have to convert the ppm value in percentage then we need to divide it by 1,000,000/100 = 10,000)

Sample standard deviation = S = 17.74252 ppm = 17.74252 / 10000 = 0.001774%

For the second sample, we are given

Sample size = n = 6

Sample mean = Xbar = 0.0882%

Sample standard deviation = S = 0.730%

Now, we have to find the confidence interval for difference between two population means.

Confidence level = 95%

degrees of freedom = n1 + n2 - 2 = 6 + 5 - 2 = 9

Critical t value = 2.2622 (by using t-table or excel)

Confidence interval = (X1bar – X2bar) ± t*sqrt[(S1^2/n1)+(S2^2/n2)]

Confidence interval = (0.086122 – 0.0882) ± 2.2622*sqrt[(0.001774^2/5)+(0.73^2/6)]

Confidence interval = (0.086122 – 0.0882) ± 2.2622* 0.3295

Confidence interval = (0.086122 – 0.0882) ± 0.7453

Confidence interval = -0.0021 ± 0.7453

Lower limit = -0.0021 - 0.7453 = -0.7474

Upper limit = -0.0021 + 0.7453 = 0.7432

Confidence interval = (-0.7474, 0.7432)

The value ‘0’ of difference is included in the above confidence interval, so we conclude that there is no statistically significant difference in the two soil samples.

There is insufficient evidence to conclude that two soil samples significantly differ from each other.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---