a sample of swimming pool water was checked for its homogeneity of disinfectant

Question

a sample of swimming pool water was checked for its homogeneity of disinfectant level , chlorine . sample of the water were taken from the top and bottom levels of the pool.
..............,...............
top={1.50, 1.47, 1.44, 1.83, 1.39}
bottom={0.96, 1.02, 0.85, 0.93}
...............................
i) check whether there is any outlier in each of two data sets at 95% confidence level. (PLEASE SHOW CALCULATIONS)
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ii) from decision above, calculate the confidence limit for mean concentration of the 'top' data set at 95% confidence level. (PLEASE SHOW CALCULATION)
............................
iii) determine the pooled standard deviation for the above analysis. Show by using hypothesis testing whether there is any significance difference between the two result chlorine levels at 95% confidence (calc)
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thank you in advance

Explanation / Answer

Solution

Back-up Theory

95% confidence interval for the mean is {Mean x ± (s/n)(tn – 1, 0.025) where Mean X = sample mean, s = sample standard deviation, n = sample size and tn – 1, 0.025 = upper 2.5% point of t-distribution with

degrees of freedom = n – 1.

Calculations

For ‘top’ level

Given n = 5, mean X = 1.526, s = 0.5526, tn – 1, 0.025 = t4, 0.025 = 2.776,

95% confidence interval for mean is (0.7590, 2.2930)

Since none of the given values lies outside this interval, there is no outlier. ANSWER for one part of (i)

For ‘bottom’ level

Given n = 4, mean X = 0.94, s = 0.0707, tn – 1, 0.025 = t3, 0.025 = 3.182,

95% confidence interval for mean is (0.8101, 1.0699)

Since none of the given values lies outside this interval, there is no outlier. ANSWER for other part of (i)

For part (ii)

Since there are no outliers, the confidence interval remains as above.

Viz. (0.7590, 2.2930) ANSWER

For part (iii)

Pooled variance = {(n1 - 1)s1^2 + (n2 - 1)s2^2}/(n1 + n2 - 2), where n1. n2 are respective sample sizes (5, 4) and s1 and s2 are respective sample standard deviations.

Thus, Pooled variance = 1.2362/7 = 0.1766. Hence,

Pooled standard deviation = 0.1766 = 0.4204 ANSWER

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