An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores? Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.1 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1 2 3 4 5 6 7
Score on first SAT 500 380 560 430 450 360 560
Score on second SAT 540 470 580 450 480 400 600

Required:
a. State the null and alternative hypotheses for the test.
b. Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.
c. Compute the value of the test statistic. Round your answer to three decimal places.
d. Determine the decision rule for rejecting the null hypothesis. Round the numerical portion of your answer to three decimal places.
e. Make the decision for the hypothesis test.

Since the calculated value of t = -0.215  falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05.

e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.

Step-by-step explanation:

Student                        1      2      3       4      5         6          7

Score on first SAT    500   380 560   430  450    360    560

Score on second SAT 540 470 580  450    480    400    600

Difference d                -40  -90   -20    -20    -30     -40     -40     ∑ -280

d²                               1600  8100 400 400   900   1600  1600   ∑14600

a. Let the hypotheses be

H0:  ud= 0      against the claim Ha: ud ≠0

The degrees of freedom = n-1= 7-1= 6

The significance level is 0.05

The test statistic is

t= d`/sd/√n

The critical region is ║t║≤ t (0.025,6) = ±2.447

d`= ∑di/n= -280/7= -4

Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]

= 1/6[14600-(-4)²/7] = [14600-2.2857/6]= 2432.952

b. Sd= 49.3249= 49.325

Therefore

c. t= d`/ sd/√n

t=   -4/ 49.325/√7

t= -4/18.6435 = -0.2145= -0.215

d. Since the calculated value of t = -0.215  falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05

e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1