chapter 6 solution

CHAPTER 6

6.1 a) Data arranged in bins with width 2mm

6.2 a) Data arranged in bins with width 2in

6.1

b)

b)

Bin(in)

Number of Measurements

48.1-50.0 050.1-52.0 052.1-54.0 054.1-56.0 056.1-58.0 058.1-60.0 860.1-62.0 262.1-64.0 064.1-66.0 068.1-70.0 0

Bin(cm)

Number of Measurements

48.9-49.09 149.1-49.29 249.3-49.49 249.5-49.69 049.4-49.89 149.9-50.09 150.1-50.29 250.3-50.49 050.5-50.69 1

6.3

Bin No.80-84.9 0 85-89.9 1 90-94.9 2 95-99.9 0

100-104.9 2 105-109.9 2 110-114.9 3 115-119.9 1 120-124.9 0

6.4Bin No.

8-8.49 08.5-8.99 39-9.49 3

9.5-9.99 410-10.49 2

10.5-10.99 0

6.3

6.5 Using the data from problem 6.1:Mean:

Median:

Arranging data in ascending order:48.9, 49.2, 49.2, 49.3, 49.3, 49.8, 49.9, 50.1, 50.2, 50.5

Standard Deviation:

Modes:

6.4

6.6 Using the data from problem 6.2:Mean:

Median:

Arranging data in ascending order:58.8, 59.1, 59.2, 59.3, 59.3, 59.8, 59.8, 60.0, 60.3, 60.4

Standard Deviation:

Modes:

6.5

6.7

Median:Arranging data in ascending order89, 92, 94, 100, 104, 106, 108, 110, 110, 114, 115, 120

Standard Deviation:

Mode:

6.6

6.8

Median:Arranging data in ascending order8.7, 8.8, 8.9, 9.0, 9.3, 9.4, 9.5, 9.5, 9.8, 9.9, 10.0, 10.2

Standard Deviation:

1bar

Mode:

6.9 Probability of having a 6 and a 3 in tossing two fair dice:

There are two ways of getting a 6 and a 3 - 6,3 and 3,6

6.10 Probability of having a 4 and a 2 in tossing two fair dice:

There are two ways of getting a 4 and a 2 - 4,2 and 2,4

6.11 Probability of an undergraduate electrical engineering student to be a woman:

6.7

6.12 Probability of an undergraduate biology student to be a woman:

6.13 Probability of all three components being defective:

6.14 Probability of all three components being defective:

6.15 Binomial distribution

all 5 > 12 oz.; p=0.99, n=5, r=5

all 5 < 12 oz.; p=0.99, n=5, r=0


all 5 > 8 oz.; p=0.98, n=5, r=5

all 5 < 12 oz.; p=0.98, n=5, r=0


all 6 > 3000 hours; p=0.9, n=6, r = 6

6.8


all 6 > 3600 hours; p=0.95, n=6, r = 6


Success is failure before 1000 hours. We want probability of 1 or 2 failures.

p=0.2, n=2, r=1 and r=2

The probability of 1 or 2 is then P(1)+P(2) = 0.36

6.9

6.20 Probability distribution function:

(a) f(x) satisfies the requirement of a probability distribution function because:

(b) Expected (mean) value of x:

(c) Variance of population:

6.10

6.21 Probability distribution function:

(a) f(x) satisfies the requirement of a probability distribution function because:

(b) Expected (mean) value of x:

(c) Variance of population:

6.11

6.22 Probability of the following cases of problem 6.20:

(a) for

(b) for

6.23 Cumulative distribution of random variable x:

F(-2) = 0 , F(0) = 0.229 and F(3) = 1

6.12

6.24 Binomial distribution can be used because of the satisfactory/ unsatisfactory outcome of the process.

In this case: p=0.95n=4

(a) All four parts be satisfactory:

(b) For at least two parts to be satisfactory, we should calculate the probability that 2,3 and 4 parts be satisfactory:

Probability of having at least two satisfactory parts:

6.25 We want the probability that at most 2 computers will fail. The is the probability that 0, 1 or 2 failures. Define “success” as a computer failure. Then p=0.1. Then we want the probability that 0, 1 or 2 will fail.

P(2 or less) = .121577 + 0.27017 + 0.28518 = 0.677

6.13

6.26 Define success = failure. P = 0.1We want the probability that 2, 3, 4, or 5 will fail.

P(2-5) = P(2)+P(3)+P(4)+P(5)for example:

P(2-5) = 0.2852+.1901+.098+.032 = 0.597

6.27 (a) For all 6 parts to be satisfactory, using binomial distribution:

(b) For at least two parts to be satisfactory, we should find the sum of probabilities for 2,3,4,5 and 6 parts to be satisfactory:

6.14

6.28 Probability of one or more power failure = 0.05Probability of no power failure (success) = 0.95Binomial distribution will be used here:

a- No power failure in three months: n=3, r=3

b- Exactly one month with power failure in four months: n=4, r=3

c- At least one power failure in the nest five months: n=5

Chance of at least on power failure in 5 months.

6.29 Binomial distribution can be used:success = a failure n=16 r=0 p=0.01

6.15


(a) p=0.05, n=100, r = 2, 5, 10

Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However 100!/(2!98!) can be rewritten 100x99x98!/(2!98!) =100x99/2!, which can be readily computed.

(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99

This must be solved by trial and error for p. Using a spreadsheet, the answer is p=0.0155


(a) p=0.03, n=100, r = 1, 4, 15

Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However 100!/(1!99!) can be rewritten 100x99x98!/(1!99!) =100/1!, which can be readily computed.

(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99

This must be solved by trial and error for p. Using a spreadsheet, the answer is p=0.0155

6.32 We are looking at the probability that more than 175 passengers will show up. This can be solved as a binomial distribution problem. Consider success that a passenger shows up, so p=0.95. We then want more than 175 successes out of 180 trials.

6.16

P(r>175) = P(176)+ P(177)+ P(178)+ P(179)+ P(180) = 0.03174+0.01363+0.004363+0.0009263+0.00009778 = 0.0508

6.17

6.33 We are looking for the probability that there will 5 or less defective components. Consider success to be a defective component then p = 0.05, n=55 and r = 0,1,2,3,4,5

r=0:

r=1:

r=2:

r=3:

r=4:

r=5:

So the probability of 5 or less defective components is 0.0595+0.1725+0.2449+0.2277+0.1558+0.0836 = 0.944. This is the probability that there are 50 or more good components.

6.34 This is a Poisson distribution problem. = 40/8 = 5 visits/hour. The probability of more than 5 visits in an hour is 1 – [P(5)+P(4)+ P(3)+P(2)+P(1)+P(0)].

So P(x>5) = =0.384. The probability that there will be more than 5 visits in an hour is 0.384

6.35 This can be solved as either a binomial distribution (p=0.001, n=4, r=1) or as a Poisson distribution. For the poisson distibution, the expected occurrence () will be 4/1000 = 0.004. We are then looking for the probability of x = 1.

.

6.36 This is a Poisson distribution. The expected number of crashes, , is (1/3)5=1.666. Then,

6.37 This is a Poisson distribution problem although it can be done as a binomial distribution. The expected frequency () of defects will be (1/50)x10 = 0.2 for the typical 10 m2 kitchen. The probability of 1 or more defects is 1-P(0).

P(x1) = 1-0.818 = 0.182. If solved as a binomial distribution , p= 0.02, n=10. For r = 0,

. So P(x1) = 1-0.818 = 0.182

6.38 Poisson distribution. The fact that a bulb has failed does not affect the probability of other failures. We need to know what the probability of two or more failures during the day. The probability of 2 or more is:P(x2) = 1 – P(0) – P(1). = 2.

P(x2) = 1 – 0.1355 – 0.2707 = 0.5938

6.18

6.39 Poisson distribution. The expected number of failures in 50 calls will be (5/100)x50 = 2.5.(a) The probability of exactly 5 failures is:

(b) P(5 or less) = P(0)+P(1)+P(2)+P(3)+P(4)+p(5) Similar to part (a)P(5 or less) = 0.08208+0.2052+0.2565+0.2137+0.1336+.0668 = 0.958(c) P(more than 5) = 1 – P(5 or less) = 1 – 0.958 =0.042

6.40 Poisson distribution. The average value of customers in that 1 hour is 20 = .(a) (b)P(20 to 25) = P(20)+ P(21)+ P(22)+ P(23)+ P(24)+ P(25)=0.0888+0.0846+0.0769+0.0669+.0557 = 0.417558(c)P(10 or less) = = 0.010812(d) P(x>10) = 1 – P(x10) = 1 – 0.010812 = 0.989

6.41 Poisson distribution. =3 for 1 sheet(a) (b) P(0) = 0.049787(c) =3/6 = 0.5 for a single board. (d) P(more than 1) = 1 – P(1) – P(0) = 1 – 0.3033 - 0.6065 = 0.0902

6.42 For P(1or more) = 0.01, P(0) = 0.99 = Solving for we get = 0.01 defects per board or 0.06 defects per sheet.

6.19

6.43The area from 100 to 100.5 is 0.2. From Table 6.3, z = 0.52

Probability of error greater than 0.75 Volts:

From the normal distribution curve(Table 6.3) for z = 0.7812 P(z) = 0.2826

for error greater than 100.75 or less than 99.25 (1000.75) we will have:

6.44The area from 110 to 110.5 is 0.25. From Table 6.3, z = 0.67

Probability of error greater than 1 Volts:

From the normal distribution curve(Table 6.3) for z = 1.33 P(z) = 0.4082

for error greater than 111 or less than 109 (1101) we will have:

6.20

6.45a)

From Table 6.3 area = 0.1915 100(0.1915)(2)= 38 readings within 0.5 cm

b)

From Table 6.3 area = 0.4772 100(0.4772)(2)

= 95 readings within 2 cm

c)

From Table 6.3 area = 0.5 100(0.5)(2)= 100 readings within 5 cm

d)

From Table 6.3 area = 0.0398100(0.0398)(2)= 8 readings within 10 cm

6.21

6.46a)


b)

From Table 6.3 area = 0.4772 20(0.4772)(2)

= 19 readings within 2 cm

c)


d)

From Table 6.3 area = 0.520(0.5)(2)= 20 readings within 10 cm

6.22

6.47 (a) The average is 71.3, the median is 70 and S = 12.62

(b) The grades according to the criterion will be:score grade

95 A86 B+83 B+79 B79 B78 B75 B-70 C+70 C+68 C+63 C63 C55 C55 C-50 D

(c) There are 15 students and the division will be:

grade no. of students

A 0.342A- 0.66B+ 1.3785B 2.247B- 2.8725C+ 2.8725C 2.247C- 1.3785D 0.66F 0.342Of course, the number of students with each grade is an integer. If we round off, we will only get 14 students total so some judgement is required for the additional student.

6.23

6.48

a) From Table 6.3 for area0.4; z = 1.28

x = 9,360 hrs

b)

6.49 Normal distribution problem.

(a) The probability of the weight being less than 7.9 is the area under the curve to the left of x = 7.9. For this value of x, . From the nomral distribution table, for z = 3.0, the area under the curve (from the mean to the value of interest) is 0.4987. The area to the left of x = 7.9 is thus 0.5 – 0.4987 = 0.0013. So 0.13% of the cans will be rejected.(b) For this case, . For this value of z, the area under the normal distribution curve from the mean to x=7.9 is 0.4772. So the area to the left of 7.9 is 0.5 – 0.4772 = 0.0228. 2.28% of the cans will be rejected.(c) for this case, The table does not give a value for z this high but for 4.9, it is given as 0.5. So the area to the left of 7.9 is 0.5 – 0.5 = 0. For this z value the number of rejected cans is negligible. (actually 3x10-5%.

6.50 (a) A 2 tolerance means that values greater than z = 2 or less than z =-2. For z = 2.0, the value from the normal distribution table is 0.4772 . The right hand tail area (for z>2) is 0.5 – 0.4772 = 0.0228 The left hand tail has the same area. The probability of rejection is then 2x0.0228 = 0.0456 or 4.56%.(b) This part is best solved using the binomial distribution. For 2 rejections, we define success as a rejection. Then for 2 rejections, r = 2, n=20 and p = 0.056. The probability

6.24

of 2 rejected parts is then 0.1705. Similarly, for r = 4 and r = 10 we get 0.0099 and 4.5x10-9 respectively.

6.51

% rejection = 100(1-0.8664) = 13.36%

To reduce the rejection rate to 3%1-P = 0.03 P = 0.97 z(for P/2 = 0.485) = 2.17

This means that the manufacturer has to cut the columns close to the standard size.

6.52

a)

b)

From Table 6.3, A = 0.4987 From Table 6.3

c) From Table 6.3,

6.25

6.53

a) Using Table 6.3

b)

From Table 6.3, A = 0.4938 From Table 6.3

c) From Table 6.3,

6.54

a) = 5000From table 6.3 for area = 0.40 z 1.28

x = 43,600 miles

6.26

b) area = 0.4772

area = 0.5

0.5-0.4772 = 0.0228

(100,000 tires)(0.0228) = 2280 tires fail between 60,000 to 70,000 miles

c)

No tires expected to have life less than 20,000

d) Major assumption: Life span of tire follow normal distribution.

6.55 = 160 hrsFor A = 0.4, z 1.28

Recommend replacing bulbs after 3395 hours.

6.56

6.57a Confidence level: 95%1- = 0.95 = 0.05

0.5 - /2 = 0.475 z = 1.96 (Table 6.3)

6.27

6.57b For a large sample, n>30,

For 90, 95, and 99% confidence levels, x/2 = 1.64, 1.96, and 2.58 respectively.For 90% confidence level, the confidence interval is 1.64x0.2/(40)1/2 = 0.052 oz. For 95% and 99% confidence levels, the intervals are 0.061 oz and 0.082 oz.

6.58 In this case the sample size is small, less than 30, so we must use the t-distribution. The interval on the mean is given by: . The sample size is 20 so the degrees of freedom is 19. From the t-distribution table, the values of t for = 19 and the confidence levels of 90%, 95% and 99% (/2 = 0.05, 0.025 and 0.005) are 1.729, 2.093, and 2.861. For 90% confidence level, the confidence interval is 1.729x0.2/(20)1/2 = 0.0773 oz. For 95% and 99% confidence levels, the intervals are 0.0936 oz and 0.128 oz.

6.59 (a) The mean is 16.042 oz. and the sample standard deviation is 0.079 oz. The standard deviation of the mean is 0.07941/(12)1/2 = 0.0229 oz.(b) This is a t-distribution problem with = 12-1 = 11 and /2 = 0.025. From the t-distribution table, t has a value of 2.201. The confidence interval on the mean is then t/2S/(n)1/2 = 2.201x0.0229 = 0.0504 oz.

6.28

6.60 Find 95% confidence interval on the mean

1- = 0.95 = 0.050.5-/2 = 0.475 z = 1.96 (From Table 6.3)

6.29

6.61

Find the 95% confidence interval on the mean of the life of the VCRs.

a)

Confidence interval on the mean is 107 hours

b) For 95% confidence interval on the mean of 50 hrs:

Assuming S =150 hrs will remain the same (in reality should be recalculated and it may change), using Table 6.6 we find n by trial and error.

n

15 2.145 8320 2.093 7025 2.064 6236 2 50

So 36 systems should be tested. is an approximate value for n > 30, in which case the t-distribution

approaches normal distribution.

6.30

6.62 For this problem, we need to find a one-sided confidence interval of the form

t-distribution(a) = 41.25x106, S = 0.30x106, n=10. For 99% confidence level, = 0.01 and = 10 – 1 = 9 from Table t = 2.821. The tolerance interval is then:

41.0x106 falls into this interval so the manufacturer cannot be confident that the average strength exceeds this limit.

(b) If we change n to 20, then = 20 – 1 =19 and t = 2.539.

Now the manufacturer can be confident the mean exceeds 41.0x106 psi.

6.31

6.63

Cut confidence interval by one half > 0.57/2 = 0.285Assuming that n > 30, and S to remain the same:

20 more motors should be tested.

6.32

6.64

Cut confidence interval by one half > 0.42/2 = 0.21Assuming that n > 30, and S to remain the same:

22 more motors should be tested.

6.33

6.65 Given: Confidence level = 99% Find: NSheight = 2.44 in.Shand = 0.34 in.

Soln.: Assuming N > 30

For height: the confidence interval:

For sleeve length:

In order to satisfy both conditions, 40 members should be chosen.

6.66

Find the 95% confidence interval.

6.34

6.67 The distribution governing the time interval between the arrival successive cars is give to be

f(t,λ) = λe-λt Where

(a)

(b)

(c) In order to have only two arrivals in 10 minutes, if the interval between the first

two arrivals is t, mins, the interval between the next arrival should be greater than (10-t1

)

Thus,

P[no more than 2 arrivals] = P[0≤t1

≤10 and t2

≥(10-t1

)]

6.68 We are given that TBF has an exponential distribution with μ = 450 hours. Therefore

(a)

(b)

(c)

6.35

6.69 Let the pollution level, X, be in ppm. If we define y = log(X), then y is N(µ,σ), where μ = 1.9031 and σ = 1.3010

Note: Here we have used y = log(X) rather than y = ln(X)

(a)

We now convert this problem to a standard normal distribution,

Using the standard normal distribution in Table 6.3,

(b)

(c) If the mean pollutant level is reduced to 40 ppm, then the new value of μ = log(40 = 1.6021

6.37

6.70 Let X be the time taken to fix the software bug. X has a mean of 200 mins, a standard deviation of 30 mins. If y = log(X) is N(μ,σ) with μ = log(200) = 2.3010, σ = log(30) = 1.4771

(a)

We now convert this problem to a standard normal distribution,

(b)

(c)

6.71

Chi-squared distribution.S = 0,002 in., n = 10, = n – 1 = 9. 95% confidence level: = 1-0.95 = 0.05./2 = 0.025, 1-/2 = 0.975. From Table 6.7, 2

/2 = 19.023, 21-/2 = 2.704.

or

6.72

Chi squared distribution.S = 5500, n = 8, = 7. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.2

/2 = 20.278, 21-/2 = 0.9893.

or

6.38

6.73

Chi squared distribution.S = 10, n = 12, = 11. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.2

/2 = 26.757, 21-/2 = 2.6032.

or We cannot be 99% confident that the standard deviation is less 15 mA.

6.74

There is an error in the problem. The required variance should be 0.0004 mm2. (not m2).

(a) Chi squared distribution.S = 0.01, n = 10, = 9. = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975.2

/2 = 19.023, 21-/2 = 2.7004.

or

(b) The maximum of the confidence interval is less than the desired variance so the part is acceptable.

(c) S = 0.01, n = 5, = 4. = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975.2

/2 = 11.143, 21-/2 = 0.4844.

The upper end of the confidence interval exceeds the allowable so the part is not acceptable.

6.39

6.75 There is an error in the problem. The required variance should be 0.0004 mm2. (not m2).

(a) Chi squared distribution.S = 0.01, n = 10, = 9. = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.2

/2 = 23.589, 21-/2 = 1.7349.

(b) The maximum of the confidence interval is greater than the desired variance so the part is not acceptable.

6.76 There is an error in the problem. The required variance should be 0.0004 mm2. (not m2).

(a) Chi squared distribution.S = 0.01, = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.

We need to determine n so that the maximum of the confidence interval is 0.0004. We are told that n = 10 was too small so n must be larger. The following calculations were performed on a spreadsheet program.

It can be concluded that 13 measurements will make the part acceptable.

6.40

6.77 Data from problem one arranged in ascending order:48.9, 49.2, 49.2, 49.3 ,49.3, 49.8, 49.9, 50.1, 50.2, 50.5

From the Table 6.8:n = 10 = 1.798

S = (0.530)(1.798) = 0.95294Since S > deviations, then:No data should be required.

6.78 From problem 6.4:

n = 12 = 1.829 (Table 6.8)

Neither 1 or 2 exceeds S so No points rejected

6.41

6.79

x y

20 1.02 -33.5714 1127.04 -1.5271 2.3320 51.266930 1.53 -23.5714 555.611 -1.0171 1.0345 23.974540 2.05 -13.5714 184.183 -0.4971 0.25711 6.746350 2.55 -3.5714 12.7549 0.0029

8.4110-

6

-0.01036

60 3.07 6.4286 41.3269 0.5229 0.27342 3.361575 3.56 21.4286 459.185 1.0129 1.0260 21.7050

100 4.05 46.4286 2155.61 1.5029 2.2587 69.77754535.71 7.1717 176.8214

This value of rxy which is close to 1 shows a strong linear relationship.

6.42

6.80 (a)

xi yi (xi)2 (yi)

2 xiyi

20 1.02 400 1.0404 20.430 1.53 900 2.3409 45.940 2.05 1600 4.2025 82.050 2.55 2500 6.5025 127.560 3.07 3600 9.4249 184.275 3.56 5625 12.6736 267.0100 4.05 10000 16.4025 405

(b)

6.43

6.81(a) xi(t) yi(t)

0.0000 4.98 -0.2500 0.0625 3.6683 13.4567 -0.9171 0.1000 1.84 -0.1500 0.0225 0.5283 0.2791 -0.0793 0.2000 0.68 -0.0500 0.0025 -0.6317 0.3990 0.0316 0.3000 0.25 0.0500 0.0025 -1.0617 1.1271 -0.0531 0.4000 0.09 0.1500 0.0225 -1.2217 1.4925 -0.1833 0.5000 0.03 0.2500 0.0625 -1.2817 1.6427 -0.3204

=0.1750 =18.3971 =-1.5215

Note: The relationship between Voltage and time is not linear!

(b)xi(t) yi(lnV)

0.0000 1.6054 -0.2500 0.0625 2.5173 6.3368 -0.6293 0.1000 0.6098 -0.1500 0.0225 1.5216 2.3154 -0.2282 0.2000 -0.3857 -0.0500 0.0025 0.5262 0.2769 -0.0263 0.3000 -1.3863 0.0500 0.0025 -0.4744 0.2251 -0.0237 0.4000 -2.4079 0.1500 0.0225 -1.4961 2.2382 -0.2244 0.5000 -3.5066 0.2500 0.0625 -2.5947 6.7324 -0.6487

=0.1750 =18.124 =-1.780

, showing a closer linear relationship between

ln(V) and than between V and t.

6.45

6.82

For the best fit line, the equ. is reading=1.0525weight-1.1661For the forced zero, reading=0.9987weight.

6.46

y = 1.0525x - 1.1661

y = 0.9987x

-10

0

10

20

30

40

0 10 20 30 40

weight - lb

read

ing

6.83

6.84 For the data given in Problem 6.79, we first compute the following quantities, using x = T and y = V

We can now compute the slope, a, and the intercept, b, as

Therefore the best fit is V = 0.0390T + 0.4590

(a) The two-sided 95% confidence interval implies α/2 = 0.025 so that t

α/2,n-2

= t

0.025,5

=

2.571, the standard error,

The 95% two-sided confidence interval for a is

6.47

or (0.0302 ≤ a ≤ 0.0479)

Similarly the 95% confidence interval for the intercept, b, is

(b) When T = 70o

C, V = aT + b = 3.1929V

The 95% two-sided prediction interval for the above V is

where

x* = 70 and y* = 3.1929V

Therefore the 95% prediction interval for V at T = 70o

C is

(c) The upper limit of V at T = 70o

C at 95% confidence level involves a one-sided

prediction interval with α = 0.05.

At α = 0.05, tα,n-2

= t0.05,5

= 2.015

Thus, the upper limit will be

6.85 Based on the given data, we compute the following quantities, where x = ToF and y = σ in ksi.

6.48

(a) The independent variable is the temperature, T

o

F, and the dependent variable is

the tensile strength, σ (ksi). The coefficients of the linear regression model are

Therefore the linear regression model for the data is

σ = -0.0372T + 54.099

(b) When the temperature is T = 670

o

F, the expected σ is

σ

exp

= -0.0372x670+54.099 = 29.2056 ksi.

For a 90% two-sided confidence interval, α/2 = 0.05, n = 12.

From the tables, t

α/2,n-2

= t

0.05,10

= 1.812

The confidence interval for strength, σ, at T = 670

o

F is

where

x* = 670 and

6.49

Therefore, 90% confidence interval for σ at T = 670

o

F is

(c) The expected loss of strength for each 100

o

F increase is

Δσ = a x 100 = -0.0372 x 100 = -3.72 ksi

The 95% confidence interval for this change in strength is 100 x (95% confidence

interval for a)

The 95% confidence interval for the slope parameter, a, is

Therefore, 95% confidence interval for change in strength for a 100

o

F change is =

(-4.34, -3.09) ksi

(d) At T = 550

o

F, the expected tensile strength is

σ = a x 550 + b = -0.0372 + 550 + 54.09 = 33.641 ksi

The one-sided 95% confidence level for the strength is

where

6.50

x* = 550, α = 0.05, n = 12

From the tables, t

α,n-2

= t

0.05,10

= 1.812

Then,

= 33.641-0.0051 = 33.6359 ksi

Therefore, theh minimum strength at T = 550

o

F at a 95% confidence level is

σ

min

= 33.6359 ksi

6.51

6.86a) Best fit line for t and V

xi (t) xi2 yi (V) yi

2 xiyi

0.0 0 4.98 24.800 00.1 0.01 1.84 3.3856 0.1840.2 0.04 0.68 0.4624 0.1360.3 0.09 0.25 0.0625 0.0750.4 0.16 0.09 0.0081 0.0360.5 0.25 0.03 0.0009 0.015

The best fit line is therefore: y = -8.694x + 3.485

Best fit line for t and lnV

xi xi2 yi yi

2 xiyi

0.0 0 1.6054 2.5774 00.1 0.01 0.60977 0.37181 0.0609770.2 0.04 -0.38566 0.14874 -0.0771320.3 0.09 -1.3863 1.9218 -0.415890.4 0.16 -2.4079 5.7982 -0.963160.5 0.25 3.5066 12.2959 -1.7533

6.52

Best fit line is therefore: y = -10.175x + 1.6319

b) Standard error of the estimate for T vs V

Standard error of the estimate for T vs ln(V)

The standard error for (T vs ln(V)) is much less than for (t vs V) therefore there is less data scatter around this best fit curve.

6.53

6.87

The ln(Q) vs ln(DelP) is clearly the better fit than the straight day (Q vs. DelP). The fit with log10 is different but is equivalent to the ln fit and is just as good. The standardized residuals were computed for the ln regression analysis. The value of the standardized residual at DelP = 0.19 is 1.702 and this point is suspect although it is not clearly an outlier.

6.54

y = 12.179x + 1.5506

00.5

11.5

22.5

33.5

44.5

5

0 0.05 0.1 0.15 0.2 0.25 0.3

DelP- psi

Q-c

fm

y = 0.5017x + 2.1965

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

-4 -3 -2 -1 0

ln(DelP)

ln(Q

)

-2

-1

0

1

2

0 0.05 0.1 0.15 0.2 0.25 0.3

DelP

e/Sx

.y

y = 0.5017x + 0.9539

00.10.20.30.40.50.60.7

-1.5 -1 -0.5 0

log10(DelP)

log1

0(Q

)

6.88

6.55

2nd order fit

y = -0.0236x2 + 0.9779x - 2.597401234

4 5 6 7 8

V - m/s

Q -

l/s

6.89

This data is non-linear and the 2nd order fit is superior. Note the points at the ends –the linear fit is above them and the 2nd order passes through them.

6.56

linear fit

y = 0.695x - 1.796

01234

4 5 6 7 8

V - m/s

Q -

l/s

6.90

6.91(a)

(b) Due to the S shape of the curve, at least a 3rd order curve will be required.

(c) The 4th order curve fit is shown.

6.57

y = 0.0373x2 + 0.11x + 0.0698

0

0.1

0.2

0.3

0.4

0 0.5 1 1.5

VoltageVe

loci

ty

6.92

Let L1 be the length of the shelves and L2 be the length between sides of the cabinet. The clearance between the two is = L2 – L1. We want to be 99% confident that is greater than equal to zero.

Assume large sample size for the standard deviations so we can use the normal distribution. We want to be sure that 99% of the shelves will not be too long. Using Table 6.3, For 99% confidence level, A = 0.49 and z =2.33

.This value of z corresponds to

The longest the nominal length the shelves can be is 24 – 0.156 = 23.844 in long. Note: We did not concern ourselves with how short the shelves can be.

6.58

chapter 6 solution

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