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

Increasing the sample size decreases the margin of error, E. c.

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 are 90% confident that the amount of time it takes a customer to go through taco bell’s drive thru is between 161.5 and 164.7 seconds.

27

Two reccomendations for increasing the precision of the interval based on the “hours Worked” survey in Problem 23 are:

  1. Increase the size of the sample
  2. Make the level of condience smaller in order to narrow the confidence interval

29

  1. A large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive BAC because the sample has to be large in order for the distribution of the sample mean to meet the requirements of a normal distribution.

  2. This, along with the fact that the data were obstained using a simple random sample, satisfies the requirements for constructing a confidence interval because the sample size is less than 5% of the population

  3. This means that were are 90% confident that the mean BAC in fatal crashes where the driver had a positive BAC is between 0.165 and 0.169 g/dL.

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. This is possible because there is a chance that the correct mean for this distribution may not have fallen within the confidence interval. However, this is unlikely.

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 communicates that we are 99% confident that the mean number of books read by Americans in the past year was between 12.05 and 14.75.

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 communicates that we are 95% confident that the mean incubation period of patients with SARS is between 1.08 and 8.12.

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

Increasing the sample size makes the width of the interval decrease.

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 level of confidence makes the width of the interval increase.

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

This means that we can be 95% confident that the population standard deviation of the prices of 4 GB flash memory cards at online retailers is between 1.61 dollars and 4.28 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 means that we can be 90% confident that the population standard deviation of repair costs of a low-impact bumper crash on a minicar or microcar is between 849.69 dollars and 1655.35 dollars.