Question

Math SAT Scores (Raw Data, Software Required):
Suppose the national mean SAT score in mathematics is 520. The scores from a random sample of 40 graduates from Stevens High are given in the table below. Use this data to test the claim that the mean SAT score for all Stevens High graduates is the same as the national average. Test this claim at the 0.10 significance level.

(a) What type of test is this? This is a left-tailed test. This is a two-tailed test.     This is a right-tailed test. (b) What is the test statistic? Round your answer to 2 decimal places. tx= (c) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value = (d) What is the conclusion regarding the null hypothesis? reject H0 fail to reject H0     (e) Choose the appropriate concluding statement. There is enough data to justify rejection of the claim that the mean math SAT score for Stevens High graduates is the same as the national average. There is not enough data to justify rejection of the claim that the mean math SAT score for Stevens High graduates is the same as the national average.    We have proven that the mean math SAT score for Stevens High graduates is the same as the national average.

DATA ( n = 40 ) MATH SAT Scores    504 523 542 506 454 412 513 487 473 546 461 510 529 430 561 461 520 514 586 555 549 516 485 524 563 504 506 526 548 563 512 513 513 510 519 546 558 448 577 492

Answer 1

Given: = 520,

From the data: = 513.98, s = 39.466, n = 40, = 0.10

The Hypothesis:

H0: = 520

Ha: 520

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(a) Option 2: This is a 2 tailed test

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(b) The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

t observed = -0.97

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(c) The p Value:    The p value for t = -0.97, for degrees of freedom (df) = n-1 = 39, is; p value = 0.3380

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(d) Conclusion: Since p value is > : Fail to Reject H0.

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(e) Option 2: There is not enough data to justify rejection of the claim that the mean SAT math scores for Stevens High graduates is the same as the national average.

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#	X	Mean	(x - mean)2
1	504	513.98	99.60
2	523	513.98	81.36
3	542	513.98	785.12
4	506	513.98	63.68
5	454	513.98	3597.60
6	412	513.98	10399.92
7	513	513.98	0.96
8	487	513.98	727.92
9	473	513.98	1679.36
10	546	513.98	1025.28
11	461	513.98	2806.88
12	510	513.98	15.84
13	529	513.98	225.60
14	430	513.98	7052.64
15	561	513.98	2210.88
16	461	513.98	2806.88
17	520	513.98	36.24
18	514	513.98	0.00
19	586	513.98	5186.88
20	555	513.98	1682.64
21	549	513.98	1226.40
22	516	513.98	4.08
23	485	513.98	839.84
24	524	513.98	100.40
25	563	513.98	2402.96
26	504	513.98	99.600
27	506	513.98	63.680
28	526	513.98	144.4804
29	548	513.98	1157.3604
30	563	513.98	2402.9604
31	512	513.98	3.9204
32	513	513.98	0.9604
33	513	513.98	0.9604
34	510	513.98	15.8404
35	519	513.98	25.2004
36	546	513.98	1025.2804
37	558	513.98	1937.7604
38	448	513.98	4353.3604
39	577	513.98	3971.5204
40	492	513.98	483.1204

n	40
Sum	20559
Average	513.98
SS	60744.976
Variance = SS/n-1	1557.563487
Std Dev	39.4660