In: Statistics and Probability
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We use t-distribution with small samples.
1)
The sheriff’s office samples in LEMAS (N = 827) reported that their agencies require new recruits to complete a mean of 599 hours of academy training (s=226). Construct a 99% confidence interval around this mean and interpret the interval in words.
Given sample size n = 827
sample mean = 599
sample std dev s = 226
Confidence level = 99%
level of significance alpha = 100 - confidence level = 1%
(1-)*100 Confidence Interval = ()
t0.005,826 = 2.582
99% CI = () = (578.709 , 619.291)
We are 99% confident that mean hours of acedamy training will lie in between (578.709 , 619.291).
2)
In the PPCS, respondents who say that they have been stopped by police while driving vehicles are asked to report the reason why they were stopped and the total length of time that the stop took. Among female drivers stopped for illegal use of a cellphone while driving (N=42), stops lasted a mean of 9.00 minutes (s=5.48). Construct a 95% confidence interval around this sample mean and interpret the interval in words.
Given sample size n = 42
sample mean = 9
sample std dev s = 5.48
Confidence level = 95%
level of significance alpha = 100 - confidence level = 1%
(1-)*100 Confidence Interval = ()
t0.025,41 = 2.02
99% CI = () = (7.292, 10.708)
We are 95% confident that illegal use of cell phone mean time of stops for female drivers lasted will lie in between (7.292, 10.708).
3.) Respondents to the GSS (N=1,166) worked a mean of 40.27 hours per week (s=15.54). Construct a 99% confidence interval around this sample value and interpret the interval in words.
Given sample size n = 1166
sample mean = 40.27
sample std dev s =15.54
Confidence level = 99%
level of significance alpha = 100 - confidence level = 1%
(1-)*100 Confidence Interval = ()
t0.005,1165 = 2.58
99% CI = () = (39.096, 41.444)
We are 99% confident that mean working time of GSS workers will lie in between (39.096, 41.444).
4.)In a random sample of prisons (N=30) from the COJ, the mean number of inmates per security-staff member was 4.34 (s=3.94). Construct a 95% confidence interval around this sample value and interpret the interval in words.
Given sample size n = 30
sample mean = 4.34
sample std dev s = 3.94
Confidence level = 95%
level of significance alpha = 100 - confidence level = 1%
(1-)*100 Confidence Interval = ()
t0.025,29 = 2.045
99% CI = () = (2.869, 5.811)
We are 95% confident that mean number of inmates per security-staff member will lie in between (2.869, 5.811).