9.2

21

n = 35
xbar = 18.4
s = 4.5
t_critical = qt(.975, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(16.85,19.95))
## [1] 16.85 19.95
n = 50
xbar = 18.4
s = 4.5
t_critical = qt(.975, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(17.12,19.68))
## [1] 17.12 19.68
n = 35
xbar = 18.4
s = 4.5
t_critical = qt(.995, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(16.32,20.48))
## [1] 16.32 20.48

It appears the margin of error increases.

  1. The population must be normal.

23

  1. A confidence interval is NOT a probability interval.

  2. Correct

  3. A confidence interval is NOT a census.

  4. We are making a statement about the population parameter of the whole country, NOT just Idaho.

25

We are 90% confident that the mean drive through service time of Taco Bell restaurants is between 161.5 and 164.7

27

Increase the sample size and decrease the level of confidence

29

  1. A large sample size is needed because it gives less margin of error and because the data is not normally distributed, this is possible

  2. This gives a large sample size and distribution was normal because data was applied to a large mean

n = 51
xbar = .167
s = .01
t_critical = qt(.95, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(0.165,0.1693))
## [1] 0.1650 0.1693
  1. It is possible that the mean BAC is not captured in the confidence interval, but it is not likely.

31

n = 1006
xbar = 13.4
s = 16.6
t_critical = qt(.995, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(12.1,14.7))
## [1] 12.1 14.7

We are 99% confident that the mean number of books hardcover and paperback read all or part of the way by Americans in the past year is 12.1 and 14.7.

33

n = 81
xbar = 4.6
s = 15.9
t_critical = qt(.975, n - 1)
lower = xbar - t_critical*s/sqrt(n)
upper = xbar + t_critical*s/sqrt(n)

(answer = c(1.05,8.12))
## [1] 1.05 8.12

We are 95% confident that the mean incubation period of the SARS virus is between 8.12 days and 1.05 days.

9.3

5

n = 20
(small_value = qchisq(.05, n-1)) 
## [1] 10.11701
(large_value = qchisq(.95, n-1))
## [1] 30.14353

7

n = 23
(small_value = qchisq(.01, n-1))
## [1] 9.542492
(large_value = qchisq(.99, n-1))
## [1] 40.28936

9

n = 20
ssquared = 12.6
small_value = qchisq(.05, n-1)
large_value = qchisq(.95, n-1)

lower = (n-1)*ssquared/large_value
upper = (n-1)*ssquared/small_value

(answer = c(7.942,23.66))
## [1]  7.942 23.660
n = 30
ssquared = 12.6
small_value = qchisq(.05, n-1)
large_value = qchisq(.95, n-1)

lower = (n-1)*ssquared/large_value
upper = (n-1)*ssquared/small_value

(answer = c(8.59,20.63))
## [1]  8.59 20.63

Increasingg the sample size decreases the width of the sample.

n = 20
ssquared = 12.6
small_value = qchisq(.01, n-1)
large_value = qchisq(.99, n-1)

lower = (n-1)*ssquared/large_value
upper = (n-1)*ssquared/small_value

(answer = c(6.61,31.36))
## [1]  6.61 31.36

Increasing the level of confidence increases the width

11

n = 10
ssquared = (2.343)^2
small_value = qchisq(.025, n-1)
large_value = qchisq(.975, n-1)

lower = (n-1)*ssquared/large_value
upper = (n-1)*ssquared/small_value

(answer = c(1.61,4.23))
## [1] 1.61 4.23

We are 95% confident that the population standard deviation price for a 4GB flash memory card is between 1.61 and 4.23

13

n = 14
ssquared = (1114.412)^2
small_value = qchisq(.05, n-1)
large_value = qchisq(.95, n-1)

lower = (n-1)*ssquared/large_value
upper = (n-1)*ssquared/small_value

(answer = c(849.69,1655))
## [1]  849.69 1655.00