Question

2 (a)

The chloride content of water samples has been determined a very large number of times using a particular method, and the standard deviation found to be 7 ppm. Further analysis of a particular sample gave experimental values of 350 ppm (i) for a single determination, (ii) for the mean of two replicates and (iii) for the mean of four replicates. At the 95% probability level, z = 1.96. Find the confidence limits in (i), (ii) and (iii).

2 (b) The same chloride analysis as in Question 2a, but using a new method for which the standard deviation was not known, gave the following replicate results: mean = 351.67 ppm and estimated standard deviation = 6.66 ppm. At the 95% probability level, z = 1.96. Find the confidence limits.

Answer 1

2a. The confidence interval is given by the expression

CI = x̅ + Z*S/√n

where x̅ is the mean of the measurements, Z = 1.96 for the 95% Confidence Limit, S is the standard deviation of the mean and n is the number of measurements.

(i) Here, x̅ = 350 ppm, S = 7 ppm and n = 1.

Plug in values and get

CI = (350 ppm) + (1.96)*(7 ppm)/√(1)

= (350 ppm) + (13.72 ppm)

The CI is from (350 - 13.72) ppm = 336.28 ppm to (350 + 13.72) ppm = 363.72 ppm (ans).

(ii) Here, n = 2.

CI = (350 ppm) + (1.96)*(7 ppm)/√(2)

= (350 ppm) + (13.72 ppm)/(1.414)

= (350 ppm) + (9.70 ppm)

The CI is from (350 - 9.70) ppm = 340.30 ppm to (350 + 9.70) ppm = 359.70 ppm (ans).

(iii) Here, n = 4.

CI = (350 ppm) + (1.96)*(7 ppm)/√(4)

= (350 ppm) + (13.72 ppm)/(2)

= (350 ppm) + (6.86 ppm)

The CI is from (350 - 6.86) ppm = 343.14 ppm to (350 + 6.86) ppm = 356.86 ppm (ans).

2b. Here, x̅ = 351.67 ppm, S = 6.66 ppm and n = 1.

Plug in values and get

CI = (351.67 ppm) + (1.96)*(6.66 ppm)/√(1)

= (351.67 ppm) + (13.05 ppm)

The CI is from (351.67 - 13.05) ppm = 336.62 ppm to (351.67 + 13.05) ppm = 364.72 ppm (ans).