(a) Company-A claims that bulbs are superior to those of its main competitor, co

Question

(a) Company-A claims that bulbs are superior to those of its main competitor, company-B. The text showed that a sample of n_1 = 40 of company-A's bulbs had a mean lifetime of 647 hours of continuous use with a standard deviation of 27 hours while a sample of n_2 = 40 bulbs made by company B had a mean lifetime of 638 hours of continuous use with a standard deviation of 31 hours. Does this test indicate that the bulbs produced by company-A have 25 hour of longer lifetime than competitor B? [Use significance level (alpha) of the 0.01]. What is the P-value? (b) The lapping process which is used to grind certain silicon wafers to the proper thickness is acceptable only if the population standard deviation (sigma) of dice cut from the wafers, is at most 0.5 micron. A testing was carried out for 15 dice cut from such wafers and it was found that standard deviation of sample = 0.64 micron. Does this test indicate that original claim of sigma = 0.5 micron is true? (Use sigma = 0.05 level of significance)

Explanation / Answer

Q2a

Let X = Life (hours) of bulbs produced by Company A and Y = Life (hours) of bulbs produced by Company B.

Then, we assume, X ~ N(µ1, 12), Y ~ N(µ2, 22) and 12 = 22 = 2, say, but unknown,

Claim: Bulbs produced by Company A have 25 hour longer life than the bulbs produced by Company B.

Hypotheses:

Null H0: µ1 - µ2 = 25 [claim] Vs Alternative HA: µ1 - µ2 < 25

Test statistic:

t = [(Xbar - Ybar) – (µ1 - µ2)]/[s(2/n)], where Xbar and Ybar are respectively the sample means of X and Y; s = {s12 + s22}/2; s1 and s2 being sample standard deviations of X and Y.

and n = common sample size.

Calculations:

s = {272 + 312}/2 = 845 = 29.068

t = {(647 - 638) – 25}/{29.068 x (2/40)} = - 16/6.500 = - 2.462

So,tcal = - 2.462

Distribution and p-value

Under H0, t ~ t2n – 2 i.e., t-distribution with (2n - 2) degrees of freedom.

p-value = P(t2n – 2 < - 2.462) = P(t78 < - 2.462) = 0.008

Decision Criterion (Rejection Region)

Reject H0, if p-value < .(0.01 given)

Decision:

Since p-value < , H0 is rejected.

Conclusion:

There is not sufficient evidence to support the claim that the bulbs produced by Company A have 25 hour longer life than the bulbs produced by Company B. DONE

Q2(b)

Let X = thickness of silicon wafers in micron

Then, X ~ N(µ, 2)

Claim: Population standard deviation is 0.5 micron.

Hypotheses:

Null H0: 2 = 20   = 0.25 Vs Alternative HA: 2 > 0.25

Test statistic:

2 = (n - 1)s2/ 20 where

20 (given) = 0.25

s = sample standard deviation = 0.64

n = sample size = 15.

2 = (14 x 0.642)/0.25 = 7.2132

So, 2cal = 7.2132

Distribution and p-value

Under H0, 2 ~ 2n - 1

So, p-value = P(2n – 1 > 7.2132) = P(214 > 7.2132) = 0.926 [using Excel Function]

Decision Criterion (Rejection Region)

Reject H0, if p-value < (0.05 given).

Decision:

Since p-value > , H0 is accepted.

Conclusion:

There is sufficient evidence to support the claim that the population standard deviation is 0.5 micron. DONE

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