Home Work 3

Chapter 8

Md. Tarique Hasan Khan

ID: fl8214

8-12. The diameter of holes for a cable harness is known to have a normal distribution with 0.01 inch. A random sample of size 10 yields an average diameter of 1.5045 inch. Find a 99% two-sided confidence interval on the mean hole diameter

Answer:

We eill have to find a 99% two-sided confidence interval on the mean hole diameter.

99% Two-sided CI on the diameter cable harness holes: where

=1.5045 ,

s = 0.01 ,

n=10 and

Z0.005 = 2.58

Now,

- Z0.005 σ /√n <= µ <= + Z0.005 σ /√n

1.504-2.58(0.01)/ √10 <= µ <= 1.504+2.58(0.01)/ √10

1.4963 <= µ <= 1.5127

So, this is the confidence interval.

8-23.

Determine the t-percentile that is required to constructeach of the following two-sided confidence intervals:(a) Confidence level = 95%, degrees of freedom = 12(b) Confidence level = 95%, degrees of freedom = 24(c) Confidence level = 99%, degrees of freedom = 13(d) Confidence level = 99.9%, degrees of freedom = 15

Answer:

So we will have to find out t0.025,12 , t0.025,24, t0.025,13, t0.025,15

By using the t-distribution table:a.) t0.025,12 =2.179b.) t0.025,24, =2.064c.) t0.025,13=3.012d.) t0.025,15=4.073

8-28. An Izod impact test was performed on 20 specimens of PVC pipe. The sample mean is = 1.25 and the sample standard deviation is s = 0.25. Find a 99% lower confidence bound on Izod impact strength.

Answer:

We will have to find the lower confidence bound on Izod Impact strength.99% lower confidence bound on mean Izod impact strengthn=20,

= 1.25,S=0.25,t0.01,19 = 2.539

- t0.01,19(s/√n) <= µ1.25-2.539(0.25/√20) <= µ1.108<= µSo, this is the lower confidence bound on mean Izod impact strength.

8-34.Answer:

The data looks like normally distributed on the basis of examination of the normal probability plot below. So, there is proof to support that the solar energy is normally distributed.

By using Minitab, Probability Plot of Solar Energy has been drawn.

95% confidence interval on mean solar energy consumedn=16,

= 65.58,S=4.225,t0.025,15 = 2.131

- t0.025,15(s/√n) <= µ <= + t0.025,15(s/√n)65.58-2.131(4.225/√16) <= µ <= 65.58+2.131(4.225/√16)63.329<= µ <= 67.831So this is the 95% confidence interval for this problem.

8-51.Answer:Here,95% confidence interval for σAgain,

Solar Energy

Probability Plot of Solar Energy

n=15,s=0.00831χ2

α/2,n-1 = χ20.025,14=26.12 and χ2

1- α,n-1 = χ20.95,14=6.53

σ2 ≤ 14*(0.00831)2/(6.53)σ2 ≤0.000148σ ≤0.0122

By using Minitab, Probability Plot of Gauge Capability has been drawn.

The data do not appear to be normally distributed in the above normal probability plot . For that reason, the 95% confidence interval for σ is not valid.

8-57Answer:a.95% two-sided confidence interval on the true proportion of rats that would show underweight.

Gauge Capability

Probability Plot of Gauge Capability

pH = (12/30)=0.4n=30zα/2=1.96

pH - zα/2 √[ pH (1- pH )/n] <= p <= pH + zα/2 √[ pH (1- pH )/n]0.4-1.96 √ [0.4(1-0.4)/30] <= p <= 0.4+1.96 √ [0.4(1-0.4)/30]0.225 <= p <= 0.575

b.

E=0.02α=0.05zα/2=z0.025=1.96 andpH =0.4 as the initial estimate of p,n=( z α/2/E)2 pH (1- pH )= (1.96/0.02)20.4(1-0.4) =2304.96we can say approximately n= 2305

c.

E=0.02α=0.05zα/2=z0.025=1.96 at least 95% confidencen=( z α/2/E)2 (0.25) = (1.96/0.02)2(0.25)=2401So this should be the sample size.

8-88Answer:With σ = 8, the 95% confidence interval on the mean has length of at most 5; the error is then E = 2.5.

a.n=(z0.025/2.5)2 82

=(1.96/2.5)2 64=39.34I recommend sample size n=40

b.

n=(z0.025/2.5)2 62

=(1.96/2.5)2 36=22.13Now sample size, n=23

Here because of decreasing of standard deviation, when all the other values held constant, the sample size necessary to maintain the acceptable level of confidence and the length of the interval, decreases.