Stati sti cs - Portage Online Summer 2018
MATH 110 MODULE 6 EXAM.| PORTAGE
LEARNING
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
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
n=60 x =¯ 350 s=90
This study source was downloaded by 100000834306259 from CourseHero.com on 04-20-2022 02:35:55 GMT -05:00
https://www.coursehero.com/file/32391144/Module-6-Examdocx/
, Stati sti cs - Portage Online Summer 2018
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
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
This study source was downloaded by 100000834306259 from CourseHero.com on 04-20-2022 02:35:55 GMT -05:00
https://www.coursehero.com/file/32391144/Module-6-Examdocx/
MATH 110 MODULE 6 EXAM.| PORTAGE
LEARNING
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
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
n=60 x =¯ 350 s=90
This study source was downloaded by 100000834306259 from CourseHero.com on 04-20-2022 02:35:55 GMT -05:00
https://www.coursehero.com/file/32391144/Module-6-Examdocx/
, Stati sti cs - Portage Online Summer 2018
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
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
This study source was downloaded by 100000834306259 from CourseHero.com on 04-20-2022 02:35:55 GMT -05:00
https://www.coursehero.com/file/32391144/Module-6-Examdocx/
