The life in hours of an Electronic sensor is known to be approximately normally

Question

The life in hours of an Electronic sensor is known to be approximately normally distributed, with standard deviation sigma = 20 hours. A random sample of 10 sensors resulted in the following data: 500, 550, 560, 575, 525, 505, 510, 540, 550, 545 (a) Is there evidence to support the claim that mean life exceeds 520 hours? Use a fixed-level test with alpha=0.05. (b) What is the P-value for this test? Conclude the same of part (a)? (c) What is the Beta-value for this test if the true mean life is 535 hours? (d) What sample size would be required to ensure that Beta does not exceed 0.10 if the true mean life is 540 hours? (e) Construct a 95% one-sided lower CI on the mean life. (f) Use the CI found in part (e) to test the hypothesis. (g) show how to get the same result for part a) and b) using Minitab

Explanation / Answer

let X denotes life in hours of an electric sensor.

X is assumed to follow a normal distribution with unknown mean u and sd=sigma=20

we have a random sample of size n=10 from X with values

500,550,560,575,525,505,510,540,550,545

a) we want to test if the mean lifetime is greater than 520hours

so H0:u=520 vs H1:u>520

the test statistic is given as T=(Xbar-520)*sqrt(n)/20 whihc under H0 follows a N(0,1) distribution.

where Xbar is the sample mean

alpha=level of significance=0.05

H0 is rejected iff t>tao0.05  where t is the observed value of T and tao0.05 is the upper 0.05 point of N(0,1) distribution.

now Xbar=536 so t=(536-520)*sqrt(10)/20=2.5298

and tao0.05=1.64 so t>tao0.05

hence H0 is rejected and the conclusion is that the mean life exceeds 520 hours [answer]

b) since it is a right tailed test. hence the p avlue of the test is p=P[T>2.5298] where T~N(0,1)

so p=0.005706378<alpha

hence H0 is rejected. hence conslusion is same as part a

c) beta value=P[type 2 error]=P[accepting H0|H0 is false]

=P[(Xbar-520)*sqrt(n)/20<1.64 | Xbar~N(540,(20/sqrt(10))2)

=P[(Xbar-540+540-520)*sqrt(10)/20<1.64]

=P[(Xbar-540)*sqrt(10)/20<1.64-(540-520)*sqrt(10)*20]

=P[Z<-1.52] where Z~N(0,1)

=0.06425549 [answer]

d) let the required sample size be k.

then we must have P[(Xbar-520)*sqrt(k)/20<1.64]<=0.10 Xbar~N(540,(20/sqrt(k))2)

or, P[(Xbar-540+540-520)*sqrt(k)/20<1.64]<=0.10

or, P[(Xbar-540)*sqrt(k)/20<1.64-(540-520)*sqrt(k)/20]<=0.10

or, P[Z<1.64-sqrt(k)]<=0.10=P[Z<-1.281552]

so 1.64-sqrt(k)=-1.281552

or, k=(1.64+1.281552)2=8.5354~9

hence the required sample size is 9 [answer]

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