answer.
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
Veseljchak [2.6K]
1 year ago
7

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60

0 to 2400.
Suppose that PrepIt! is a company that offers classes to help students prepare for the SAT exam. In their ad, PrepIt! claims to produce "statistically significant" increases in SAT scores. This claim comes from a study in which 427 PrepIt! students took the SAT before and after PrepIt! classes. These students are compared to 2,733 students who took the SAT twice, without any type of formal preparation between tries.

We also conduct a hypothesis test with this data and find that students who retake the SAT without PrepIt! also do significantly better (p-value < 0.0001). So now we want to determine if PrepIt! students improve more than students who retake the SAT without going through the PrepIt! program. In a hypothesis test, the difference in sample mean improvement ("PrepIt! gain" minus "control gain") gives a p-value of 0.004. A 90% confidence interval based on this sample difference is 3.0 to 13.0.

What can we conclude?

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.
Mathematics
1 answer:
kodGreya [7K]1 year ago
5 0

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

n_1 represent the sample after the preparation

n_2 represent the sample without preparation  

\bar x_1 =678 represent the mean sample after preparation

\bar x_2 =1837 represent the mean sample without preparation

s_1 =197 represent the sample deviation after preparation

s_2 =328 represent the sample deviation without preparation

\alpha=0.1 represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})} (1)  

The point of estimate for \mu_1 -\mu_2

The appropiate degrees of freedom are df=n_1+ n_2 -2

Since the Confidence is 0.90 or 90%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}  

After replace in the formula for the confidence interval we got this:

3.0 < \mu_1 -\mu_2

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

You might be interested in
Ronelle asked 100 students to choose their favorite subject from among mathematics, history, and art. She found that 22 students
gayaneshka [121]
Number of students that chose art = 100-(22+26) = 100- 48 = 52

so percentage of students who chose art = 52%
6 0
1 year ago
Read 2 more answers
The list shows the area in square feet of each apartment available for rent in a building. 565, 961, 867, 517, 627, 714, 517, 72
gladu [14]
STEP 1:
put numbers in order from lowest to highest
517, 517, 565, 627, 714, 728, 867, 961

STEP 2:
find the difference between the lowest and highest numbers

961 - 517= 444 ft^2 range

ANSWER: The range of the areas is 444 ft^2.

Hope this helps! :)
8 0
2 years ago
Read 2 more answers
Find an equation of the line containing the centers of the two circles whose equations are given below.
Anna35 [415]

Answer:

<h2><em>3y+x = -5</em></h2>

Step-by-step explanation:

The general equation of a circle is expressed as x²+y²+2gx+2fy+c = 0 with centre at C (-g, -f).

Given the equation of the circles x²+y²−2x+4y+1  =0  and x²+y²+4x+2y+4  =0, to  get the centre of both circles,<em> we will compare both equations with the general form of the equation above as shown;</em>

For the circle with equation x²+y²−2x+4y+1  =0:

2gx = -2x

2g = -2

Divide both sides by 2:

2g/2 = -2/2

g = -1

Also, 2fy = 4y

2f = 4

f = 2

The centre of the circle is (-(-1), -2) = (1, -2)

For the circle with equation x²+y²+4x+2y+4  =0:

2gx = 4x

2g = 4

Divide both sides by 2:

2g/2 = 4/2

g = 2

Also, 2fy = 2y

2f = 2

f = 1

The centre of the circle is (-2, -1)

Next is to find the equation of a line containing the two centres (1, -2) and (-2.-1).

The standard equation of a line is expressed as y = mx+c where;

m is the slope

c is the intercept

Slope m = Δy/Δx = y₂-y₁/x₂-x₁

from both centres, x₁= 1, y₁= -2, x₂ = -2 and y₂ = -1

m = -1-(-2)/-2-1

m = -1+2/-3

m = -1/3

The slope of the line is -1/3

To get the intercept c, we will substitute any of the points and the slope into the equation of the line above.

Substituting the point (-2, -1) and slope of -1/3 into the equation y = mx+c

-1 = -1/3(-2)+c

-1 = 2/3+c

c = -1-2/3

c = -5/3

Finally, we will substitute m = -1/3 and c = 05/3 into the equation y = mx+c.

y = -1/3 x + (-5/3)

y = -x/3-5/3

Multiply through by 3

3y = -x-5

3y+x = -5

<em>Hence the equation of the line containing the centers of the two circles is 3y+x = -5</em>

5 0
2 years ago
Which of the following represents a pyramidal number sequence?
Juli2301 [7.4K]

Answer: the answer is C

8 0
2 years ago
Joey is building a frame for a sandbox. The sandbox is going to be a quadrilateral that has the lengths shown. A rectangle is sh
adelina 88 [10]

Answer- it’s A

Step-by-step explanation:

8 0
1 year ago
Read 2 more answers
Other questions:
  • During the three weeks how many customers came from zones 3 and 4
    5·1 answer
  • What value of x is in the solution set of 3(x – 4) ≥ 5x 2? A) -10 B) -5 C) 5 D) 10
    10·2 answers
  • A student solved the problem below by first dividing 20 by 10. What mistake did the student make? A baseball team has won 20 gam
    6·2 answers
  • The top of a table is a parallelogram with a base of 28 cm and a height of 20 cm. Patty wants to cover the table with tiles that
    12·2 answers
  • Alice has a total of 12 dimes and nickels. She has 2 more nickels than dimes. Which equation represents the given problem situat
    6·2 answers
  • The height of a cylinder is twice the radius of its base. A cylinder has a height of 2 x and a radius of x. What expression repr
    10·2 answers
  • A rectangular prism has a length of 4.2 cm, a width of 5.8 cm, and a height of 9.6 cm. A similar prism has a length of 14.7 cm,
    12·2 answers
  • HELP QUICK!
    10·1 answer
  • Drew drives his car and Rachel drives her truck in the same direction on the highway at constant speeds for 3.25 hours. During t
    15·1 answer
  • Data on average high temperatures​ (in degrees​ Fahrenheit) in July and precipitation​ (in inches) in July for 48 cities is used
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!