1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vedmedyk [2.9K]
3 years ago
6

A survey is to be conducted to determine the average driving in miles by Minnesota State University, Mankato students. The inves

tigator wants to know how large the sample should be so that he/she can be 92% confident on the estimate and the estimate is within 1.5 miles of the true average. A similar study conducted in past and it was found that the standard deviation of the students’ driving distances was 8.2 miles.
Mathematics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92

So the answer for this case would be n=92 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =1.5 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04;0;1)", and we got z_{\alpha/2}=1.75, replacing into formula (b) we got:

n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92

So the answer for this case would be n=92 rounded up to the nearest integer

You might be interested in
Find the prime factorization of the following 2 numbers 36 and 72?
Kryger [21]

Answer:

36 =  2² x 3²

72=2³×3²

Step-by-step explanation:

5 0
3 years ago
What will be the length of the battery?
kobusy [5.1K]

Answer:

:) 17

Step-by-step explanation:

68 = 4 x L

68 div by 4 = 17

the length is 17

6 0
3 years ago
Read 2 more answers
70
BaLLatris [955]

Answer:

160 is the answer please tell me if im wrong

5 0
2 years ago
How do you solve this please help
coldgirl [10]

Answer:

thanks

Step-by-step explanation:

I have been patient trying to get a job

6 0
3 years ago
The sum of six times a number and 7 is 8
sukhopar [10]
(6•x) + 7 = 8
6x + 7 = 8
6x = 1
X = 1
5 0
3 years ago
Read 2 more answers
Other questions:
  • During one year, there were 89,790 births in a large metropolitan area. If
    5·1 answer
  • John can reach 60 centimeters above the top of his head. If John is 2 meters tall, how high can he reach?
    5·1 answer
  • There were 150 sheets of paper in a 5 subject notebook how many sheets paper are in a 6 subject notebook
    12·1 answer
  • A grocery store sells 6 bottles of water for 4$ and 18 bottles of water for 10. Is the cost of the water proportional to the num
    12·1 answer
  • Helppppppppppp plzzz ​
    6·1 answer
  • I need help Estimate 81/4 but into a number
    5·2 answers
  • Please help me with this please
    7·1 answer
  • What is the vertex of the parabola? y+1=−14(x−2)2 Enter your answer in the boxes.
    15·1 answer
  • 3. The bill at a restaurant
    11·2 answers
  • What is the value of x in the solution to this system of equations?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!