Homework 7 Solutions

p. 731

3.

The setup should be a random drawing of 900 from a box of 50,000 tickets. Of the 50,000, 10,000 should be 1's and 40,000 should be zero.

• c) Do the math. SEBOX is not equal to $20,000.
• e) P( 19% IS LESS THAN PERCENT IS LESS THAN 20% ) = P(-1 < Z < 1) = 68% From EBOX=20%, SEBOX=1%

7.

• a) True. Expected values can be calculated.
• b) False. It is exactly 25%.
• c) True. The percentage of draws will vary.
• d) False. See c).
• e) True. It is our population.
• f) False. Our estimate of the population parameter will vary.

11.

• a) Observed=357, Expected=340
• b) Observed=71%, Expected=68%

p. 392

4.

• a) 36.1% +/- 1.2%
• b) 95.2% +/- .55%

5.

False. The sample percentage is likely to be near the population percentage.

7.

Sure.

10.

SESUM=SQRT(N)*SDBOX=10*SDBOX
SEAVG=SDBOX/SQRT(N)=SDBOX/10
ESUM=N*EBOX
EAVG=EBOX

P( SUM > 710)=P(Z > (710 - 100EBOX)/(10SDBOX))=P(Z > (71-10*EBOX)/SDBOX))
P( AVG > 7.1)=P(Z > (7.1 - EBOX)/(SDBOX/10))=P(Z > (71-10*EBOX)/SDBOX))

Equal.

p.425

1.

• a) P(80 < AVG < 120) = P(-20 < AVG - 100 < 20)= P(-20 < Z < 20)=1
• b) P(99 < AVG < 101) = P(-1 < AVG < 1)= 68%

3.

• a) 8.7 miles, SDAVG=.9 miles
• b) 8.7 +/- 1.8

8.

SD=6500
AVGEN=3700
SEAVGEN=325
N=400

• a) True.
• b) True.
• c) False.
• d) False.
• e) False. N is large, we are using averages, which should approach a normal curve.

p. 497

1.

• a) True. See definition, p.502 #4.
• b) False. See definition, on .501 #1.

2.

• a)
  H_o: p=18/38
  H_a: p > 18/38
  
• b) P( Z > (.50-.47)/.01) = P( Z > 3) =~ 0
• c) Yes.