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

This means that on average, customers will wait between 161.5 and 164.7 seconds 90% of the time they go to the Taco Bell drive thru.

27

To increase the precision of the interval of the poll, one could increase the sample size or increase the level of confidence.

29

  1. A large sample size is needed to construct a confidence interval for the mean BAC because a smaller sample size would possible give distorted or inaccurate results.

  2. In order to contruct a confidence interval, the sample size has to be large or normally distrbuted.

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
  1. It is not possible for the mean of all drivers involved in fatal accidents to be less than the legal intoxication level because since the crashes were fatal, it is more likely that the drivers were more likely to be over the legal amount than the legal level.

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

This interval means that 99% of the mean of the population will fall between 12.04 and 14.75 books read in the past year.

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

This interval means that 95% of the mean of the population was incubated for 1.08 to 8.11 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(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

As the sample size increases, the width of the confidence interval also increases because the sample gives a more accurate representation of the population.

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

As the level of confidence increases, the width of the confidence interval also increases.

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

The interval means that 95% of the mean of the prices of the memory cards will be between 1.61 and 4.27 dollars.

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

This interval means that we are 90% confident that the mean price to repair a bumper crash will be between 849.69 and 1655.35.