Portage Learning 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 population and large sample size
n=60 xx =350 s=90 z=1.282
xx -z s <u< xx +z s
√n √n
350-1.282 90 <u<350+1.282 90
√60 √60
335.105<u<364.895
-1.0 points
Instructor Comments
You have a small math error in your calculation.
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
n=60 x ¯=350 s=90
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 275 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 2670 calories per day with a standard deviation of 330 calories
per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the
school.
Case 3: Finite population
N=275 n=40 xx =2670 s=330 z=1.96
xx -z s √N-n<u< xx +z s √N-n
√n √N-1 √n √N-1
2670-1.96 330 √ 275-40<u< 2670+1.96 330 √ 275-40
√40 √275-1 √40 √ 275-1
2575.291<u<2764.709
Instructor Comments
Excellent!
Answer Key
A certain school has 275 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 2670 calories per day with a standard deviation of 330 calories
per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the
school.
The population is finite. So, we should use Case 3: Finite population.
Use:
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 population and large sample size
n=60 xx =350 s=90 z=1.282
xx -z s <u< xx +z s
√n √n
350-1.282 90 <u<350+1.282 90
√60 √60
335.105<u<364.895
-1.0 points
Instructor Comments
You have a small math error in your calculation.
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
n=60 x ¯=350 s=90
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 275 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 2670 calories per day with a standard deviation of 330 calories
per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the
school.
Case 3: Finite population
N=275 n=40 xx =2670 s=330 z=1.96
xx -z s √N-n<u< xx +z s √N-n
√n √N-1 √n √N-1
2670-1.96 330 √ 275-40<u< 2670+1.96 330 √ 275-40
√40 √275-1 √40 √ 275-1
2575.291<u<2764.709
Instructor Comments
Excellent!
Answer Key
A certain school has 275 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 2670 calories per day with a standard deviation of 330 calories
per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the
school.
The population is finite. So, we should use Case 3: Finite population.
Use:
