Question

Given an SRM coal sample containing 41.7 ppm Fe, replicate experimental measurements yieldconcentrations of 44.0, 41.6, 45.3 and 48.7 ppm Fe giving a mean of 44.9 ppm and a standarddeviation of 2.961 ppm. What is the 95% confidence interval on the data?

a) 44.9±4.711      b) 44.9±4.1099      c)44.9±3.4836       d) 44.9±5.2254        e) other

Does the method in question #5 reproduce the known value using 90% confidence intervals?

a) Yes    b)   no

Another method was applied to the coal sample in #5 and the 4 measurements resulted in thesame mean (44.9 ppm) with a standard deviation of 3.356. Is this standard deviationsignificantly different from the first one at the 95% confidence level? what test can be used to determine?

a) yes, F-test   b)   yes, t-test    c) no, F-test    d) no, linear regression e) yes, ANO

Answer 1

Solution.

The confidence interval can be found by the formula:

x̅ ± Z_a/2 * σ/√(n)

x̅ = 44.9;

_{a = 0.95;}

σ = 2.961;

n = 4

Z0.475 = 1.96; (obtained from reference tables)

Using 90% confidence level, the confidence interval is (Z0.450 =1.645)

44.9-2.435 = 42.645 => the method in question #5 does not reproduce the known value using 90% confidence intervals.

Is this standard deviationsignificantly different from the first one at the 95% confidence level? It can be cheched performing F-test.

Square both standard deviations to get the variances: s₁ = 2.961² = 8.768; s₂ = 3.356² = 11.263; s₂/s₁ = 1.285;

Degrees of freedom: 4-1 = 3.  The F-value from the reference table is 9.277. The F-table value is greater than the calculated value, we must accept the null hypothesis - two population variances are equal.

The answer is c) no, F-test.