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(16.85,19.95))
## [1] 16.85 19.95
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(17.12,19.68))
## [1] 17.12 19.68
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(16.33,20.48))
## [1] 16.33 20.48
It appears the margin of error increases.
23
A confidence interval is NOT a probability interval.
Correct
A confidence interval is NOT a census.
We are making a statement about the population parameter of the whole country, NOT just Idaho.
25
You can say with 90% confidence that the population mean drive through service time falls between 161.5 and 164.7
27
Two ways to increase the precision of the interval is to increase the sample size or decrease the confidence level
29
Need a larger sample size for distribution to be closer to normal. also since the population isn’t normal, the larger sample size is needed to compute.
n = 25,000 sample size is sufficiently large
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(0.165,0.169))
## [1] 0.165 0.169
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(12.05,14.75))
## [1] 12.05 14.75
We can say with 99% confidence that the mean number of books Americans read last year 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(1.08,8.12))
## [1] 1.08 8.12
We can say with 95% confidence that the mean incubation period for the SARS virus falls between 1.08 and 8.12 days
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(7.94,23.66))
## [1] 7.94 23.66
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(8.59,20.63))
## [1] 8.59 20.63
Increase in sample size = decrease in interval width
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(6.61,31.36))
## [1] 6.61 31.36
Increase level of confidence = increase confidence interval width
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(1.612,4.278)))
## [1] 1.269646 2.068333
We can be 95% confident that the population standard deviation for these flash memory cards is 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(849.7,1655.3)))
## [1] 29.14961 40.68538
We can say with 90% confidence that the population standard deviation of repair costs is between 849.7 and 1655.3 dollars.