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jek_recluse [69]
3 years ago
10

An SAT coaching company claims it's course can raise SAT scores of high school students (thus, when they take it a second time a

fter being coached). Of course, students who retake the SAT without paying for coaching generally raise their scores also. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were not coached. For both the coached group and the uncoached group, the gain in score was recorded. The SAT coaching company wishes to test to see if their coaching provided better second attempts on average. What case is best
Mathematics
1 answer:
Maru [420]3 years ago
6 0

Answer:

Check the explanation

Step-by-step explanation:

(a) The appropriate test is the matched-pairs test because a student’s score on Try 1 is certainly correlated with his/her score on Try 2. Using the differences, we have xbar =  29 and s = 59.

(b) To test H0: mu=0 vs.  H1 mu > 0, we compute

t = (29-0)/((59/sqrt(427))=10.16

with df = 426. This is certainly significant, with P < 0.0005. Coached students do improve their scores on average

(a) H0: μ1 = μ2 vs. Ha: μ1 > μ2, where μ1 is the mean gain among all coached students and μ2 is the mean gain among uncoached students. H0 and Ha. Using the conservative approach, df = 426 is rounded down to df = 100 in  (t table) and we obtain 0.0025 < P < 0.005. Using software, df = 534.45 and P = 0.004. There is evidence that coached students had a greater average increase.

(b) 8 ± t*(3.0235) where t* equals 2.626 (using df = 100 with (t table) ) or 2.585 (df = 534.45 with software). This gives either 0.06 to 15.94 points, or 0.184 to 15.816 points, respectively.

(c) Increasing one’s score by 0 to 16 points is not likely to make a difference in being granted admission or scholarships from any colleges.

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When you have a negative exponent, then you have to flip the fraction (all whole numbers are over 1 so flip the 1 to the top and the 5^7 on the bottom)

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3 years ago
Isn’t 3 to the -2 power 9 right?
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Lyles bought a pack of 12 cookies. One-third of the cookies are peanut butter. How many of the cookies in the pack are peanut bu
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Two cubic meters of mercury have a mass of 27190 kilograms. What is the density of mercury?
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7 0
4 years ago
Consider the following probability distribution for stocks A and B: State Probability Return on Stock A Return on Stock B 1 0.10
jek_recluse [69]

Answer:

None of the above. The correct answer is 1.47%, 1.10%.

Step-by-step explanation:

The first thing to do is to calculate the Expected return of Stock A and Stock B.

For A;

Probability. Return.

0.1.= 0.1 × 10% = 1.00%

0.2= 0.2 × 13% = 2.60%

0.2= 0.2× 12% = 2.40%.

0.3= 0.3 × 14% = 4.20%

0.2= 0.2 × 15% = 3.00%

Total = 13.20%.

For B;

Probability. Return.

0.1= 0.1 × 8% = 0.80%

0.2= 0.2 × 7% = 1.40%

0.2.= 0.2 × 6% = 1.20%

0.3.= 0.3 × 9% =2.70%

0.2.= 0.2 × 8%= 1.60%

Total = 7.70%.

Hence, the Expected return of Stock A and B = 13.20% and 7.70%. respectively.

Now, let us find the Standard deviation of Stock A and the Standard deviation of Stock B.

For A

For individual value of A, we use the following formula;

(A - Expected return of Stock A)^2 × probability.

For instance,

(10% - 13.20%)^2 × 0.1 =0.000102.

(13% - 13.20%)^2 × 0.2= 0.000001.

(12% - 13.20%)^2 × 0.2 = 0.000029.

(14% - 13.20%)^2 × 0.3 =0.000019.

(15% - 13.20%)^2 × 0.2 = 0.000065.

Which gives us the following values;

0.000102, 0.000001, 0.000029, 0.000019, 0.000065.

The next thing to do is to find the variance (that is the addition of all the values above) and the value for the square root of variance which is the standard deviation.

√( 0.000102 + 0.000001 + 0.000029 + 0.000019 + 0.000065) = 0.000216.

Thus, variance = 0.000216, square root of variance= 0.014697(1.47%).

For B;

We follow as the one above.

(B - Expected return of Stock A)^2 × probability.

We have values as;

0.000001, 0.000010, 0.000058, 0.000051, 0.000002.

√ ( 0.000001 + 0.000010 + 0.000058 +0.000051 + 0.000002).

= 0.011( 1.10%).

8 0
3 years ago
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