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Answer Key Homework 9 (Due date: April 20, 2005, Wednesday in class)

• If the ( is not given numerically in any question use 0.05.

1. Exercise 7.25 (Hint: Notice that you are computing conservative sample size in part (b). Check the large sample conditions to use the large sample formula. If they are not satisfied, just tell me that you cannot compute them because conditions are not satisfied. Do not use the formula 7.10 in the textbook to answer.)

Answer part (a) and as a part (b) calculate the 99% confidence interval for the true proportion of the electrote aware of her position on using the state funds to pay for abortions when the sample proportion is 0.6 for 100 residents.

(a) 1-( =0.95 ( ( =0.05. (/2=0.025 ( [pic]=1.96 and w(0.1 then

n ([pic]=[pic]=384.16( 385. [pic] is used because it was asking the conservative sample size.

(b) [pic]=0.6, n=100

Since [pic] and [pic],

99% C.I for (: [pic]

[pic]= (0.4739,0.7262)

2. Exercise 7.37 (Hint: sample mean and standard deviation are given in the output for you to use it. Do not need to interpret. Only calculate)

(a) 95% confidence interval for ( is

[pic]=(0.8876,0.9634) where

[pic]=2.093

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

[pic]=(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 [pic] =(0.6331,1.2180)

3. Exercise 7.44 (Exercise 7.42, 5th edition) (Hint: you do not need to derive, really need to calculate it because formulas are given)

n=9, s=2.81 then 95% confidence interval for (2 is

[pic]

95% confidence interval for ( is (1.8981, 5.3830)

4. Exercise 7.47 (Exercise 7.45, 5th edition) (Hint: in part (a) it is simply asking the confidence interval and use the significance level 0.05. in part (b), do not use formula 7.10 again. Check the large sample conditions and use the large sample confidence interval)

(a) 95% confidence interval for ( is

[pic]=(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[pic]=13 ≥ 10 and n[pic]=35 ≥10, we have a large sample

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

5. Exercise 8.35 (Hint: Make sure to write the hypothesis, calculate the test statistics and write the decision and conclusion)

n=200, x: number passed=124, (=0.05 then [pic]

[pic] versus [pic](differs from 0.7)

Is the large sample conditions satisfied? Since np0=140 (10 and n(1-p0)=60(10, YES

Test statistics: [pic]

Then P-value=2P(z ................
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