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

It means that we are 90% confident that the the mean drive-through service time of Taco Bell is between 161.5 and 164.7

27

One recommendation would be to increase the sample size. Another would be to decrease the level of confidence, resulting in a narrower confidence interval.

29

  1. Because the distribution is not normally distribued(skewed right), we need a large sample size so the distribution of the sample mean will be aproximatelly normal

  2. Because the sample size is less than 5% of the population

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 possible, but is not likely. There is a chance that the true mean is not captured in the confidence interval.

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

The Lower bound is 12.05 and the upper bound is 14.75 books.Considering this, we can be 99% confident that the mean number of books read by americans 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

Lower bound is 1.08 and upper bound is 8.12 days. Considering this, we can be 95% confident that the mean incubation period of patiens 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

By increasing the sample size, the width of the interval decreases.

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

By increasing the level of confidence, the width of the interval 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 lower bound is 1.612 and the upper bound is 4.278.Considering this, we can be 95% confident that the population standard deviation of the prices of 4GB flash memory cards at online retailers in between 1.612 and 4.278 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

The lower bound is 849.7 and the upper bound is 1655.3. Considering this, we can be 90% confident that the population standard deviation of repair costs of a low impact bumper crash on a mini or micro car is between 849.7 and 1655.3 dollars.