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(lower,upper))
## [1] 16.8542 19.9458
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(lower,upper))
## [1] 17.12111 19.67889
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(lower,upper))
## [1] 16.32468 20.47532

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 estimate with 90% that the mean drive-through time is between 161.5 and 164.7 seconds

27

you could increase the confidence level or decrease the sample size

29

  1. A large sample is required so we can accurately obtain data and if the sample is approxamately normal it is easier to estimate the data. Larger number samples produce more precise estimates and decrease the margin of error.

  2. Because there are probably more than 25,000 fatal crashes in which the driver had a positive BAC satifies the requirements to construct a confidence interval. And with a sample of this size, the results are more likely to be normal.

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(lower,upper))
## [1] 0.1646533 0.1693467

I estimate with 90% confidence that the mean BAC of fatal crashes with a positive BAC are between 0.1642 and 0.1698

  1. No, because judging by the sample data taken in part c, the mean of data falls between 0.1642 and 0.1698

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(lower,upper))
## [1] 12.04932 14.75068

I estimate with 99% confidence that the mean number of books read by Americans during the preceding year is between 11.928 and 14.872.

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(lower,upper))
## [1] 1.084221 8.115779

I estimate with 95% confidence that the mean incubation period for the SARS virus is between 0.5643 and 8.636.

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(lower,upper))
## [1]  7.942004 23.663111
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(lower,upper))
## [1]  8.586138 20.634315

The chi-square distribution becomes more nearly symmetric

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(lower,upper))
## [1]  6.614928 31.364926

Increasing the confidence level increases the amount of space we expect our standard deviation to fall in, meaning that their is larger

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 = sqrt(c(lower,upper)))
## [1] 1.611598 4.277405

I estimate with 95% confidence that the standard deviation price of a 4GB flash memory card is between 1.6116 and 4.2774

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 = sqrt(c(lower,upper)))
## [1]  849.6926 1655.3548

I estimate with 90% confidence that the standard deviation repair cost of a low-impact bumper crash on a mini- or micro-car is between $849.69 and $1655.35