Answer Key for Homework 7 (Due Date: March 31, 2004, Wednesday in Class)

Answer Key for Homework 7 (Due date: March 31, 2004, Wednesday in class)

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Exercise 7.21 (Hint: Do not forget to check the large sample conditions and use the formula given in class if the large sample conditions are satisfied to get credit.)

95% lower confidence bound for p:

=0.2162

where

Exercise 7.23 (Hint: Notice that you are computing conservative sample size in part (b). Do not forget to check the large sample conditions and use the formula given in class for both parts if the large sample conditions are satisfied to get credit.)

=0.6487

a. Since and ,

99% C.I for p:

= (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.

Exercise 7.26 (You do not need to derive the formula since handout 7 (example 8) explains it. Just compute 95% confidence interval numerically)

By using the data,

=4.06

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

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)

Exercise 7.41 (Exercise 7.39 , 5th edition)

(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.

(b) The area on the left of –0.86 is the same as the area on the right of 0.86 with the df=20 and it is 0.20. The area on the right of 1.325 is 0.10. Then the area in the middle is 1-(0.20+0.10)=0.70.

(c) The area on the left of –1.064 is the same as the area on the right of 1.064 with the df=20 and it is 0.15. The area on the right of 1.064 is 0.15. Then the area in the middle is 1-(0.15+0.15)=0.70.

The interval in (c), it makes more sense

Exercise 7.43 (Exercise 7.41 , 5th edition)

(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

Exercise 7.45 (Exercise 7.43 , 5th edition)

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.

8. Exercise 7.47 (Exercise 7.45, 5th edition) (Do not forget to check the large sample conditions and use the formula given in class for part (b) if the large sample conditions are satisfied to get credit.)

(a) 95% confidence interval for  is

=(6.7019 , 9.4565)

(b) There are 13 values exceeding 10 out of 48 then 13/48=0.2708 is the point estimate for the proportion of all such bonds whose strength would exceed 10.

Since n=13 ≥ 10 and n=35 ≥10, we have a large sample

95% confidence interval for p is =(0.1451 , 0.3965)

9. Exercise 7.52 (Exercise 7.50, 5th edition) (In part (b), compute the 90% confidence interval not the upper confidence bound)

X: the reaction time to a stimulusn=16, =0.214, s=0.036

(a) Assume the data are normally distributed and notice that small sample size

90% CI for  is =(0.1982 , 0.2298) where =1.753

(b) 90% confidence interval for  is

=(0.0279 , 0.0517) where=24.996 and =7.261.

=(1.1490 , 0.2791)