The following data represent the pulse rates (beats per minute) of nine students enrolled in a statistics course. Treat the nine students as a population. Complete parts (a) to
(c).
Student |
PulsePulse |
|
---|---|---|
Perpectual Bempah |
80 |
|
Megan Brooks |
63 |
|
Jeff Honeycutt |
79 |
|
Clarice Jefferson |
67 |
|
Crystal Kurtenbach |
82 |
|
Janette Lantka |
85 |
|
Kevin McCarthy |
65 |
|
Tammy Ohm |
73 |
|
Kathy Wojdya |
88 |
(a) Determine the population mean pulse.
The population mean pulse is approximately
nothing
beats per minute.
(Type an integer or decimal rounded to the nearest tenth as needed.)
(b) Determine the sample mean pulse of the following two simple random samples of size 3.size 3.
Sample 1: StartSet Kevin comma Perpectual comma Kathy EndSet{Kevin, Perpectual, Kathy}
Sample 2: StartSet Megan comma Jeff comma Perpectual EndSet{Megan, Jeff, Perpectual}
The mean pulse of sample 1 is approximately
nothing
beats per minute.
(Round to the nearest tenth as needed.)
The mean pulse of sample 2, is approximately
nothing
beats per minute.
(Round to the nearest tenth as needed.)
(c) Determine if the means of samples 1 and 2 overestimate, underestimate, or are equal to the population mean.
The mean pulse rate of sample 1
▼
underestimates
overestimates
is equal to
the population mean.The mean pulse rate of sample 2
▼
underestimates
overestimates
is equal to
the population mean.
(A) Population mean = (sum of all data values)/(total number of data values)
= (80+63+79+ 67+82+85+65+73+88)/9
= 682/9
= 75.8
(B) mean of sample 1 = (sum of all three data values)/(3)
= (65+80+88)/3
= 233/3
= 77.7
and
mean of sample 2 = (sum of all three data values)/(3)
= (80+63+79)/3
= 222/3
= 74.0
(C) it is clear that the 77.7 is greater than 75.8, i.e. sample mean 1 is greater than population mean
and
it is clear that the 74.0 is less than 75.8, i.e. sample mean 2 is less than population mean
The mean pulse rate of sample 1
▼
overestimates the population mean.
The mean pulse rate of sample 2
▼
underestimates the population mean.
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