Question

A company produces industrial wiring. One batch of wiring is specified to be 2.17 centimeters (cm) thick. A company inspects the wiring in seven locations and determines that, on the average, the wiring is about 2.17 cm thick. However, the measurements vary. It is unacceptable for the variance of the wiring to be more than 0.05 cm2. The standard deviation of the seven measurements on this batch of wiring is 0.36 cm. Use α = 0.025 to determine whether the variance on the sample wiring is too great to meet specifications. Assume wiring thickness is normally distributed.

Answer 1

NULL HYPOTHESIS H0: cm^2

ALTERNATIVE HYPOTHESIS Ha: cm^2

alpha= 0.025

chi square test statistic= (n-1)*s^2/

= (7-1)*(0.36)^2/0.05

= 6*0.1296/0.05

= 15.552

Degrees of freedom= 7-1=6

The P-Value is .016372

Since Pvalue is smaller than the level of significance .

Decision: REJECT H0

Conclusion:We have sufficient evidence to conclude that the variance on the sample wiring is too great to meet specifications.