HW 9.2 & 9.3

9.2 21-33 odd
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. Flawed; A confidence interval is not a probability interval.

  2. Correct; A confidence interval describes our confidence.

  3. Flawed; A confidence interval is not a census and does not describe the individual.

  4. Flawed; We are making a statement about the population of all adults not just adults in Idaho.

25

This means we are 90% confident that the mean of drive through service is between 161.5 and 164.7

27

Increase sample size and decrease confidence to reduce interval being measured.

29

  1. A large sample sized is needed because a large sample size will make the distribution of the sample normal or very close to normal.

  2. We can construct a confidence interval with the data previously obtained because it 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. Yes it is possible because not an accurate mean may be constructed by 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 12.04932 and the upper bound is 14.75068. We can be 99% confident that the mean number of books read by Americans is 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

The lower bound is 1.084221 days and the upper bound is 8.115779 days We can be 95% confident that the mean incubation period of patients is between 1.084221 and 8.115779.

9.3 5-13 odds

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 width of the interval decreases when you increase the sample size.

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

The width of the interval increases as you increase the level of confidence.

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 1.611598 and the upper bound is 4.277405. We can be 95% confident that the population standard deviation of 4GB flash drive prices is between 1.611598 and 4.277405.

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 849.6926 and the upper bound is 1655.3548. We can be 90% confident that the population standard deviation repair costs is between 849.6926 and 1655.3548