Question

Two quality control t

echnicians measured the surface finish of a metal part, obtaining the data in

Table 4E.1. Assume that the measurements are normally distributed.

Technician 1: (1.45, 1.37,1.21,1.54,1.48,1.29,1.34)

Technican 2: ( 1.54, 1.41, 1.56, 1.37, 1.20 ,1.31, 1.27, 1.35 )

(d)

Test the hypothesis that the variances of the measurements made by the two technicians are equal.

Use What are the

practical implications if the null hypothesis is rejected?

(e)

Construct a 95% confidence interval estimate of the ratio of the variances of technician

measurement error.?

(f)

Construct a 95% confidence interval on the variance of measurement error for technician 2.?

(g)

Does the normality assumption seem reasonable for the data?

Answer 1

Test and CI for Two Variances: technician 1, technician 2

Method

Null hypothesis         σ(technician 1) / σ(technician 2) = 1
Alternative hypothesis σ(technician 1) / σ(technician 2) ≠ 1
Significance level      α = 0.05

F method was used. This method is accurate for normal data only.

Statistics

                                    95% CI for
Variable      N StDev Variance      StDevs
technician 1 7 0.115     0.013 (0.074, 0.253)
technician 2 8 0.125     0.016 (0.083, 0.254)

Ratio of standard deviations = 0.920
Ratio of variances = 0.846

95% Confidence Intervals

                            CI for
         CI for StDev      Variance
Method       Ratio           Ratio
F       (0.406, 2.195) (0.165, 4.816)

Tests

                       Test
Method DF1 DF2 Statistic P-Value
F           6    7       0.85 0.854

since p-value = 0.854 > 0.05

we fail to reject the null hypothesis

e)

95 % confidence interval for variance =

(0.165, 4.816)

f)

95 % confidence interval for variance

(0.0068, 0.0646)