9.1

25.

  1. 0.181

    1. sample was random
  1. np = 417 n(1-p) = 1889 <10
  2. sample is less than 5% = independent

  1. 0.168,0.194

  2. type answer here.

26.

  1. 0.15 population portion

  2. sample proportion .43 = 282.60

  3. 1.96

  4. between 0.402 and 0.459

27.

  1. population portion 0.519

  2. sample size is less than 5% of the pop

  3. z0.025 = 1.96 lower bound = 0.488 upper boune = 0.550

  4. yes it is possible that the pop prop is more than 60%. true prop is not captured in the confidence interval.

  5. lower bound 4.50 upper bound 5.12

28.

  1. .75

  2. = 192 >10 sample size less than 5%

  3. 99% confidence interval = 2.575

  4. upper bound 0.785 lower bound 0.715 yes it is possible that the pop prop is more than 70%

  5. upper bound 0.285 lower bound 0.215

29.

  1. 0.111

  2. 0.071,0.151

  3. 0.058, 0.164

9.2

21.

  1. lower bound 32.44 upper bound 35.96

  2. lower bound 32.73 upper bound 35.67 95% confident

  3. upper bound 36.51 lower bound 31.89

  4. cannot compute confidence interval due to the sample size being small

23.

90% confident between 161.5 and 164.7

25.

increase the number of people in survey and decrease the level fo confidence to decreese the margin of error

27.

  1. the sample distribution is normal. since its skewed right we need a large sample size to insure that the distribution is normal

  2. must be a random sample. sample must be independent. the sample size is large enough to assume normality

  3. -.1647, 0.1693

  4. 90% confidence between 0.1647 and 0.1693 g/dl. it is possible that some may be under the legal limit because we would need to do 90 intervals to capture the true pop mean.

29.

317.643 and 394.557

31.

  1. 4.8925

  2. 95% confident lies between 4.6896 and 5.0954

data <- c(4.58,5.72,5.19,4.75,5.05,5.02,4.8,4.74,4.77,4.76,4.77,4.56)
mean(data)
## [1] 4.8925
sd(data)
## [1] 0.3194064

  1. 4.6061 and 5.1789

  2. 99% confidence interval is wider than 95% because it allows you to be more confident that the pop para is within interval 33.

  3. skip

  4. skip

data2 <- c(3148,2057,1758,663,1071,2637,3345,773,743,1370)
mean(data2)
## [1] 1756.5
sd(data2)
## [1] 1007.454

  1. 95% confident that population mean cost of repair is between $1036 and $2477

  2. the cost of repair has low variability because there is only one type of car. the 95% confidence interval will be narrower