MATH 110
Exam Page 1
Suppose we have independent random samples of size n1 = 420 and n2 = 510. The
proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99%
confidence interval for the difference in the two population proportions. n1=420
n2=510 p1=.38 p2=.43 z=2.58
P1 - P2 ± z √p1 (1 - p1) + p2 (1 - p2)
n1 n2
.38 - .43 ± 2.58 √.38 (1 - .38) + .43 (1 - .43)
420 510
-.05 ± 2.58(.03227)
So the interval is ( -.1332566 .0332566 )
MATH 110 Final Exam (Latest-2025)
, MATH 110
Instructor Comments
Very good.
Answer Key
Suppose we have independent random = samples
420 and=
of510. The
1 2
size n n proportions of
success in the two samples
=.38 and= .43. Find the 99% confidence interval
1 2
are p p for the
difference in the two population
proportions.
From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the
sample sizes are
each greater than 30, so we may use
eqn. 8.2:
So, the interval is
(.-.1333,.03326).
Exam Page 2
In certain hospital, nurses are required to constantly make rounds to check in on all of
the patients. The nursing supervisor would like to know if there is a difference between
the number of rounds completed per shift by the nurses on the day shift compared to
the nurses on the night shift. So, the nursing supervisor checks the records of 61 day
shift nurses and finds that they complete an average (a mean) of 39 rounds per shift
with a standard deviation of 6.1 rounds per shift. The nursing supervisor also checks the
records of 49 night shift nurses and finds that they complete an average (a mean) of 29
rounds per shift with a standard deviation of 5.2 rounds per shift.
a) Find the 98% confidence interval for estimating the difference in the population
means (µ1 - µ2).
b) Can you be 98% confident that there is a difference in the means of the two
populations?a) z=2.33
MATH 110 Final Exam (Latest-2025)
Exam Page 1
Suppose we have independent random samples of size n1 = 420 and n2 = 510. The
proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99%
confidence interval for the difference in the two population proportions. n1=420
n2=510 p1=.38 p2=.43 z=2.58
P1 - P2 ± z √p1 (1 - p1) + p2 (1 - p2)
n1 n2
.38 - .43 ± 2.58 √.38 (1 - .38) + .43 (1 - .43)
420 510
-.05 ± 2.58(.03227)
So the interval is ( -.1332566 .0332566 )
MATH 110 Final Exam (Latest-2025)
, MATH 110
Instructor Comments
Very good.
Answer Key
Suppose we have independent random = samples
420 and=
of510. The
1 2
size n n proportions of
success in the two samples
=.38 and= .43. Find the 99% confidence interval
1 2
are p p for the
difference in the two population
proportions.
From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the
sample sizes are
each greater than 30, so we may use
eqn. 8.2:
So, the interval is
(.-.1333,.03326).
Exam Page 2
In certain hospital, nurses are required to constantly make rounds to check in on all of
the patients. The nursing supervisor would like to know if there is a difference between
the number of rounds completed per shift by the nurses on the day shift compared to
the nurses on the night shift. So, the nursing supervisor checks the records of 61 day
shift nurses and finds that they complete an average (a mean) of 39 rounds per shift
with a standard deviation of 6.1 rounds per shift. The nursing supervisor also checks the
records of 49 night shift nurses and finds that they complete an average (a mean) of 29
rounds per shift with a standard deviation of 5.2 rounds per shift.
a) Find the 98% confidence interval for estimating the difference in the population
means (µ1 - µ2).
b) Can you be 98% confident that there is a difference in the means of the two
populations?a) z=2.33
MATH 110 Final Exam (Latest-2025)
