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
mariarad [96]
2 years ago
9

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

they'd received in the mail. A 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)
Required:
Construct a 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail.
Mathematics
1 answer:
fiasKO [112]2 years ago
8 0

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that n = 603, \pi = \frac{142}{603} = 0.2355

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

You might be interested in
2 inches
Brrunno [24]

Answer: 22 square inches

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Help please!!!!!!.................
Helen [10]

Answer:

6 + 4 = 10

A. 10 inches

3 0
2 years ago
If f(n)=5^n then what is f(4)
gregori [183]
All you have to do is plug in 4 where n is

In other words n = 4

So...

F(n) = 5^n
F(4) = 5^4 or (5 x 5 x 5 x 5)
F(4) = 625
5 0
3 years ago
Explain how to use absolute value to add -8 + 12
KIM [24]

Answer:

-8+12=4 is the correct answer

Thank you

6 0
2 years ago
Read 2 more answers
Which statement is true about the graphed function?
Nady [450]

Answer:

the last option or D

Step-by-step explanation:

f(x) > 0 over the interval (-∞, -4)

i hope this helps! quizlet is also very helpful lol, have a wonderful rest of your day/night, xx, nm <3 :)

3 0
2 years ago
Read 2 more answers
Other questions:
  • You are busy baking cookies. You have baked 32 cookies and you want to make sure that each of your eight friend's cookies look t
    9·1 answer
  • The Ice-Cream Palace's sundaes include 1 cup of ice-cream, 2 ounces of chocolate syrup, and 1 ounce of sprinkles. How many ounce
    6·1 answer
  • Identify the values that create ordered pairs that are solutions to the equation 3x-5y=20
    11·1 answer
  • Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!???????????!!!!!!!!!!!!!!!!!!!!!!!
    15·1 answer
  • 1/8% is what as a decimal
    13·1 answer
  • Ali saved Rs. 36. .
    13·2 answers
  • I need help asap :):):):
    15·2 answers
  • Which of the following transformations is the only one that can change the lengths of its sides but remain similar??
    13·1 answer
  • Solve inequality 5x+3&gt;48
    7·2 answers
  • Type the correct answer in the box. If necessary, use / for the fraction bar. A shaded triangular pyramid and an inverted shaded
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!