Question

It is desired to determine the average concentration of lead in the water of a river. It uses a preliminary sample to estimate the variability. The preliminary sample indicates that the standard deviation is 0.4 Microg.

a) To estimate the lead we must first obtain the sample size that guarantees that the experiment is successful. Determine the sample size required to obtain an error in the average estimate of 0.05 g with 90% confidence.

b) Obtain the confidence interval for the standard deviation (with 99% confidence). Assume that the standard deviation of a sample of 21 units is 0.37 ug and the The standard deviation of the population is 0.43 ug.

c) Assume that the standard deviation of a sample of 21 units is 0.37 ug and the The standard deviation of the population is 0.43 ug. Calculate the probability that the standard deviation of the sample is less than 0.37 ug

Answer 1

a)

n= (z * sd /e)^2

= (1.645 * 0.4/0.05)^2

= 173.1856

=174

b)

s^2 = 0.37^2 = 0.1369

n = 21

(n- 1)s2 Xi-o/2 (n 1)s2 2 2

(0.07,0.37)

hence 99% confidence interval of standard deviation

(0.2616,0.6069)

c)

P(S < 0.37)

=P(S^2 < 0.37^2)

= P( 20s^2/0.43^2 < 20 * 0.37^2/0.43^2)

= P( $\chi^2_{20}$ < 14.808)

= =chisq.dist(14.808,20,1)

=0.2127