Answer to Question #234137 in Statistics and Probability for Jessie

Question

A new manufacturing method is supposed to increase the average life span of electronic com-
ponents, while the variance of the life span is expected to stay the same. Using the previous
manufacturing method, the average life span was 112.5 hours with a variance 12 hours. The man-
ufacturer wishes to establish the new average life span by measuring the life spans of a sample of
components manufactured using the new method.

Answer 1

(a) (a) What sample size should be used, if the manufacturer wishes to establish the new average

life span to within 1 hours, with 90% level of confidence?

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.6449.$

$$z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1$$ $$n\geq(\dfrac{z_c\sigma}{1})^2$$ $$n\geq(\dfrac{1.6449(12)}{1})^2$$ $$n\geq390$$

(b)How will the required sample size change, if the manufacturer wishes to establish the new

average life span to within 1 hours, with 95% level of confidence?

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

$$z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1$$ $$n\geq(\dfrac{z_c\sigma}{1})^2$$ $$n\geq(\dfrac{1.96(12)}{1})^2$$ $$n\geq554$$

(c) How will the required sample size change, if the manufacturer wishes to establish the new

average life span to within 1/2 hours, with 90% level of confidence?

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.6449.$

$$z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1/2$$ $$n\geq(\dfrac{z_c\sigma}{1/2})^2$$ $$n\geq(\dfrac{1.6449(12)}{1/2})^2$$ $$n\geq1559$$

The required sample size should be increased by 4 times to establish the new average life span to within 1/2 hours, with 90% level of confidence.