MATH 110 Module 6 Exam
Exam Page 1
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug
is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get
back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the
80% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
Case 1: large pop and large sample
xbar - z(s√n) < u < xbar + z(s√n)
n = 60
xbar = 350
s = 90
80% confidence interval (z) = 1.28
350 - 1.28(90√60) < u < 350 + 1.28(90√60) = 335.128, 364.872
335.13 < u < 364.87
Answer Key
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug
is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get
back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the
80% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
The drug will ultimately sold to a very large number of people. So, we may assume a very large
population. Since the sample size is greater than 30, we should use Case 1: Very large population
and very large sample size.
We are given the sample mean and sample standard deviation. So, we have
¯ 350 s=90
n=60 x =
We will use these values in the equation:
, For a 80% confidence level, we look at table 6.1 and find that z = 1.28. When we substitute these
values into our equation, we get:
When we do the arithmetic on the right and left hand side, we get:
335.13 < μ< 364.87.
Exam Page 2
A certain school has 415 male students. The school nurse would like to know how many calories the
male students consume per day. So, she samples 40 male students and finds that the mean calorie
consumption of the 40 is 2610 calories per day with a standard deviation of 560 calories per day.
Find the 80 % confidence interval for mean calorie intake of all the male students in the school.
Case 3: Finite population
x* - z(s / √n) √*(N-n)/(N-1)] < < x* + z(s / √n) √*(N-n)/(N-1)]
N = 415
n = 40
xbar = 2610
s = 560
80% confidence interval (z) = 1.28
2610 - 1.28(560 / √40) √[(415-40)/(415-1)] < u < 2610 + 1.28(560 / √40) √[(415-40)/(415-1)]
= 2502.13, 2717.87
2502.13 < u < 2717.87
Exam Page 1
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug
is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get
back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the
80% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
Case 1: large pop and large sample
xbar - z(s√n) < u < xbar + z(s√n)
n = 60
xbar = 350
s = 90
80% confidence interval (z) = 1.28
350 - 1.28(90√60) < u < 350 + 1.28(90√60) = 335.128, 364.872
335.13 < u < 364.87
Answer Key
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug
is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get
back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the
80% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
The drug will ultimately sold to a very large number of people. So, we may assume a very large
population. Since the sample size is greater than 30, we should use Case 1: Very large population
and very large sample size.
We are given the sample mean and sample standard deviation. So, we have
¯ 350 s=90
n=60 x =
We will use these values in the equation:
, For a 80% confidence level, we look at table 6.1 and find that z = 1.28. When we substitute these
values into our equation, we get:
When we do the arithmetic on the right and left hand side, we get:
335.13 < μ< 364.87.
Exam Page 2
A certain school has 415 male students. The school nurse would like to know how many calories the
male students consume per day. So, she samples 40 male students and finds that the mean calorie
consumption of the 40 is 2610 calories per day with a standard deviation of 560 calories per day.
Find the 80 % confidence interval for mean calorie intake of all the male students in the school.
Case 3: Finite population
x* - z(s / √n) √*(N-n)/(N-1)] < < x* + z(s / √n) √*(N-n)/(N-1)]
N = 415
n = 40
xbar = 2610
s = 560
80% confidence interval (z) = 1.28
2610 - 1.28(560 / √40) √[(415-40)/(415-1)] < u < 2610 + 1.28(560 / √40) √[(415-40)/(415-1)]
= 2502.13, 2717.87
2502.13 < u < 2717.87
