Question

The following are the results of 10 attempts to measure the concentration of nitrate ions (in µg/mL) in a specimen of water: 0.513 0.524 0.529 0.481 0.492 0.499 0.518 0.490 0.494 0.501 (a) Calculate a 95% confidence interval for the true concentration. (b) What assumptions have been made for the confidence interval to be valid? Are these assumptions justified? (c) ‘Quadrupling the sample size halves the width of the confidence interval’. Justify this statement.

Answer 1

(a)

From the given data, the following statistics are calculated:

n = 10

= 0.5041

s = 0.0160

SE = s/

= 0.0160/

= 0.0051

= 0.05

ndf = n - 1 = 10 - 1 = 9

From Table, critical values of t = 2.2622

Confidence Interval:

0.5041 (2.2622 X 0.0051)

= 0.5041 0.0114

= ( 0.4927 , 0.5155)

Confidence Interval:

0.4927 < <    0.5155

(b)

Assumptions:
(i) The test variable is continuous (interval or ratio level measurement)

(ii) The observations are independent

(iii) Simple Random Sample is selected

(iv) Population is approximately normal

(v) Variances of sample and population are approximately equal

(vi) There are no outlier in the data

Justification:

(i) is justified because the values are numerical

(ii), (iii), (iv) and (v) are not explicitly stated. So,they need verification.

(c)

Quadrupling the sample sizes: 4n

Width of Confidence Interval = E = Z X SE.

Since is not changed, Z is not changed.

So,

when n becomes 4n, we note that SE is halved.

So,

E =Z X SE is halved if n is quadrupled.