Answer key for Homework 7 (Due date: March 28, 2003)

  • Notice that you are working alone for this homework. If I were you, I would start doing it ASAP even though you have lots of time. It will save you from last minute frustrations.
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  • Each question worth different with the total 100 pt. for the homework.
  1. Consider a normal population distribution with the value of  known

(a) What is the confidence level for the interval ?1-2(0.0212)=0.9576 or 95.76%

(b) What is the confidence level for the interval ? 0.9788 or 97.88%

(c) What is the confidence level for the interval ? 1-0.0212=0.9788 or 97.88%

(d) What value of in the two-sided C.I. results in a confidence level 98.98%?2.57

(e) What value of in the one-sided C.I. results in a confidence level 99.43%?2.53

  1. Exercise 7.5

(a)95% C.I. for . is

(4.52,5.18)

(b)98% C.I. for . is

(4.123,4.997)

(c) (round it up) then n=55.

(d) (round it up) then n=94.

  1. Exercise 7.7

The sample size can be computed using . If you reduce the size of the width by 1/2, =4n. 4 times the sample size should be increased. If the sample size is increased by a factor of 25, . Width is reduced by factor 1/5.

  1. Exercise 7.14

(a) 95% C.I. for . is

=(88.5376,89.6624 )

(b) The sample size can be computed using 245.8624 then n=246

  1. Exercise 7.17

99% lower confidence bound for :

where

  1. Exercise 7.21

95% lower confidence bound for p:

=0.2162

where

  1. Exercise 7.23 (Hint: Notice that you are computing conservative sample size in part (b))

=0.6487

a. Since and ,

99% C.I for :

= (0.4466,0.8508)

b. 1- =0.99  =0.01. /2=0.005 =2.575 and w0.1 then

n ==663.0625664. is used because it was asking the conservative sample size.

  1. Exercise 7.26 (You do not need to derive the formula since handout 7 (example 7) explains it. Just compute it numerically)

By using the data,

=4.06

100(1-)% confidence interval for  is

  1. Exercise 7.29

(c) 1-=0.99 then =2.845

(e) =0.01 then =2.485

(f) =0.025 then =2.571

  1. Exercise 7.30

(c) 1-=0.99 then =2.947

(d) 1-=0.99 and df=5-1=4 then =4.604

  1. Exercise 7.37 (Hint: sample mean and standard deviation are given in the output for you to use it)

(a) 95% confidence interval for is

=(0.8876,0.9634) where

=2.093

(b) 95% prediction interval for a single individual randomly selected student from this population is

=(0.7520,1.0990)

(c) Tolerance interval which includes at least 99% of the cadences in the population distribution using a confidence level of 95% is =(0.6331,1.2180)

  1. Exercise 7.39

(a) the area on the left of –0.687 is the same as the area on the right of 0.687 with the df=20 and it is 0.25. The area on the right of 1.725 is 0.05. Then the area in the middle is 1-(0.25+0.05)=0.70.

  1. (6 pt.) Exercise 7.41

(a) where a is the 95th percentile of . 95% area on the left of means 5% area on the right. Since the table gives you the area on the right with 10 df, a=18.307

(c) =(1-0.025)-(1-0.975)=0.95 using df=22

  1. Exercise 7.43

Using the data sample variance of the fracture toughness is

99% C.I. for  is

=(3.5874 , 8.1439). The interval is only valid when the data comes from a normal distribution.