d. Assume o 2.0.Suppose that, instead of sampling 5items you got the same averag

Question

d. Assume o 2.0.Suppose that, instead of sampling 5items you got the same average, ? . 18.45.fom a sample of 9 hems. Test whether the assumption that ? . 20 microns is reasonable. Test at ?-OS C15) 1. The dimension for a particular feature of a printed circuit board is spedified as 20 microns plus or minus 5 microns. Historical data from our production line suggests that the current average value for our process is 18.5 microns and the standard deviation is 2.0 microns a. What percentage of our circult boards will be defective ?(15) e. lacob 1 and lacob 2 are arguing about how best to improve the process Jacob 1 believes that the best strategy is to center the process so that the true mean is actualy 20 microns and leave the standard deviation alone. Jacob 2 beleves that the best strategy is to reduce the standard deviation of the process from 2 to 1.5 microns and leave the average of the process where it is Suppose the cost is the same for each imprevement strategy. Who s correct ? Remember, are engineers and we use numbers to back up our opinions (15) b. Suppose we select 5 items at random and compute the sample average and the sample standard deviation. The results are R 18.45 and s 2.8 Test to determine if our assumption that ?·2.0 is reasonable. Test at a 'OS (15) Ho: o 2.0 Ha:0$2.0 f. Suppose we select 5 items at random and compute the sample average and the sareple standard deviation The results are 1845 and s 3.5 Test to deermine if our assumption that o 2.0 is reasonable. Test at a05 (15 c. Given your results in part b, above, test whether the assumption that u- 20 microns is reasonable. Test at ?-05 (15) s Given your resuits in part f, above, test whether the assumption that u 20 microns is reasonable. Test at a- 05 (15)

Explanation / Answer

Solution

Let X = specified characteristic of the printed circuit board.

We assume X ~ N(µ, ?2)

Given the specification for X is: 20 ± 5 microns

Given historical data suggests average of 18.5 and standard deviation of 2, we take µ = 18.5 and ? = 2. …………………………………………………………………………………(A)

Back-up Theory

Percentage of defective = 100 x P(def3ctive) = P(X is outside the specification limits) …. (B)

If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then,

Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X ? or ? t) = P[{(X - µ)/? } ? or ? {(t - µ)/? }] = P[Z ? or ? {(t - µ)/?}] .………(2)

X bar ~ N(µ, ?2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar ? or ? t) = P[Z ? or ? {(?n)(t - µ)/? }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel Function……………………..(5)

Part (a)

P(X is outside the specification limits)

= 1 - P(X is within the specification limits)

= 1 – P(15 ? X ? 25)

= 1 – {P(X ? 25) - P(X ? 15)

= 1 – P[Z ? {(25 – 18.45)/2}] + P[Z ? {(15 – 18.45)/2}] [vide (2) and (A) above]

= 1 – P(Z ? 3.275) + P(Z ? - 1.725)

= 1 – 0.9995 + 0.0423 [using Excel Function of Normal Distribution]

= 0.0923

So, vide (B) above, Percentage of defective = 9.23% ANSWER

Part (b)

H0: ?2 = 4 (i.e., ? = 2) Vs HA: ?2 > 4

Test Statistic: ?2 = (n - 1)s2/?2

= (4 x 2.82)/4 [given n = sample size = 5 and s = sample standard deviation = 2.8]

= 7.84

Since 7.84 < 9.49, the upper 5% of ?24 (Chi-square distribution with degrees of freedom = 4),

H0 is accepted. => there is not enough evidence to suggest that the population standard deviation has changed from 2. ANSWER

[The decision is based on the theory that under H0, ?2 ~ ?2n– 1 and rejection region is to reject H0 if ?2cal > ?2n– 1,?, ? being the level of significance = 0.05 or 5%]

Part (c)

H0: µ = 20 Vs HA: µ ? 20

Test Statistic: t = (?n)(Xbar - µ)/s

= (?5)(18.45 - 20)/2.8 [given n = sample size = 5, sample mean, Xbar = 18.45 and sample standard deviation, s = 2.8]

= 1.234

Since 1..234 < 2.776, the upper 2.5% of t4 (t-distribution with degrees of freedom = 4),

H0 is accepted. => there is not enough evidence to suggest that the population mean has changed from 20. ANSWER

[The decision is based on the theory that under H0, t ~ tn– 1 and rejection region is to reject H0 if tcal > tn– 1,?/2, ? being the level of significance = 0.05 or 5%]

Part (d)

H0: µ = 20 Vs HA: µ ? 20

Test Statistic: Z = (?n)(Xbar - µ)/?

= (?9)(18.45 - 20)/2 [given n = sample size = 9, sample mean, Xbar = 18.45 and population standard deviation, ? = 2]

= 2.325

Since 2..325 > 1.96, the upper 2.5% of N(0, 1),

H0 is rejected.=> there is enough evidence to suggest that the population mean has changed from 20. ANSWER

[The decision is based on the theory that under H0, Z ~ N(0, 1) and rejection region is to reject H0 if Zcal > Z?/2, ? being the level of significance = 0.05 or 5%]

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