A standardized exam consists of three parts: math, writing, and critical reading. Sample data showing the math and writing scores for a sample of 12 students who took the exam follow. Student Math Writing
1 540 474
2 432 380
3 528 463
4 574 612
5 448 414
6 502 526
7 480 430
8 499 453
9 610 615
10 572 541
11 390 335
12 593 613
(a) Use a 0.05 level of significance and test for a difference between the population mean for the math scores and the population mean for the writing scores. (Use math score − writing score.)
(b) Calculate the test statistic.

Calculated t=  0.011659

t ≥ t ( 0.025,11) = 2.201

So the null hypothesis is accepted.

Step-by-step explanation:

Student     Math      Writing            Difference                            d²

                                             math score − writing score

1                 540          474               66                                      4356

2                432         380                52                                      2704

3               528           463              65                                       4225

4               574             612             -38                                  1444

5               448            414              34                                    1156

6               502            526            -24                                   576

7               480            430              50                                 2500

8              499            453             46                                    2116

9              610             615             -5                                       25

10             572             541             31                                      961

11              390             335            55                                  3025

12             593             613          -20                                     400            

∑              6168          5856          312                                     23488                

1) We state our null and alternative hypothesis as

H0: ud= 0                   Ha: ud≠0

2) The significance level alpha is set at α= 0.05

3) The test statistic under H0 is

t= d`/ sd/√n

which has t distribution with n-1 degrees of freedom.

4) The critical region is t ≥ t ( 0.025,11) = 2.201

5) Computations

d`= ∑di/n= 312/12= 31.2

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

= 1/11[ 23488- (312)²/12]

= 1/11[ 23488- 8112] = 1/11[15376]= 5976.7272

Standard Deviation= √ 5976.7272= 77.2510= 77.25

Now t= 3.12/77.25/√12= 0.011659

The calculated t value= 0.011659 falls in the acceptance region . So we accept our null hypothesis  and reject the alternative hypothesis.

you have orange juice on wednesday. (idk why this is the answer)

step-by-step explanation:

i have no idea just need points sorry

step-by-step explanation: