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
netineya [11]
2 years ago
15

Suppose a sample of 2329 tenth graders is drawn. Of the students sampled, 489 read at or below the eighth grade level. Using the

data, estimate the proportion of tenth graders reading at or below the eighth grade level. Enter your answer as a fraction or a decimal number rounded to three decimal places. .
Using the data, construct the 98% 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level

upper end point:

lower endpoint:
Mathematics
1 answer:
Nana76 [90]2 years ago
3 0

Answer:

\hat p=\frac{489}{2329}=0.210

And the 98% confidence interval is (0.190;0.230).

Upper end point: 0.230

Lower endpoint: 0.190

Step-by-step explanation:

1) Data given and notation

n=2329 represent the random sample taken    

X represent the students who read at or below the eighth grade level

\hat p=\frac{489}{2329}=0.210 estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample  

\alpha=1-0.98=0.02 represent the significance level (no given, but is assumed)    

z_{\alpha/2} represent the quantile for the normal distribution  

p= population proportion of students who read at or below the eighth grade level

2) Solution

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.210 - 2.33 \sqrt{\frac{0.21(1-0.21)}{2329}}=0.190

0.210 + 2.33 \sqrt{\frac{0.21(1-0.21)}{2329}}=0.230

And the 98% confidence interval would be given (0.190;0.230).

We are confident that about 19% to 23% are tenth graders reading at or below the eighth grade level.  

You might be interested in
Christian has a collection of 18 shark teeth.he identified them as 6 tiger shark teeth, 8 sand shark teeth,and the rest as bull
dmitriy555 [2]

Answer:

tiger shark teeth to sand shark teeth

8 0
3 years ago
Read 2 more answers
The giraffe is 15 feet tall. she is 2 1/2 times as tall as the elephant. how tall is the elephant?
salantis [7]
The giraffe is about 5 feet tall
3 0
3 years ago
Read 2 more answers
HELP ME PLZ
N76 [4]

Answer:

b

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the relationship between the ratios? 10:14 and 30:36 Drag and drop to complete the statement. The ratios are (porportion
kari74 [83]

Answer:

Not Proportional

Step-by-step explanation:

30/10 = 3

36/14 = 2.57142857143

6 0
2 years ago
Read 2 more answers
Which expression can pair with x-y=-2 to create a consistent and dependent system?
Mandarinka [93]
-8x-3y=-2 us the answer
7 0
3 years ago
Other questions:
  • The number above the fraction the bar
    7·1 answer
  • lydia graphed triangle LMN at the coordinates L (0, 0), M(2, 2) and N(2, -1). She thinks triangle LMN is a right triangle. Is ly
    15·2 answers
  • Find the number of four-digit numbers which are not divisible by 4?
    9·1 answer
  • A^2−8ab+16b^2=81, find a–4b
    6·1 answer
  • Carmen ran 2 laps around the track in 6 minutes. What are the unit rates in laps per minute and minutes per lap?
    6·2 answers
  • Complete the factorization of 3x2 – 5x + 2.
    5·2 answers
  • 5270÷312 with remainder​
    12·2 answers
  • Can someone please help me???<br> I forgot how to do this :(
    12·1 answer
  • A particular car used 5.2 gallons of gas to go 150 miles. This
    7·1 answer
  • Helpppp PLEASEEE AHHHHHH
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!