Question

Math SAT Scores (Raw Data, Software Required): Suppose the national mean SAT score in mathematics is 510. 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 0.05 significance level.

(a) What type of test is this?

This is a left-tailed test.

This is a right-tailed test.

This is a two-tailed test.

(b) What is the test statistic? Round your answer to 2 decimal places. t x =

(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 509 528 547 511 459 417 518 492 478 551 466 515 534 435 566 466 525 519 591 560 554 521 490 529 568 509 511 531 553 568 517 518 518 515 524 551 563 453 582 497

Answer 1

Answer Date: 11/11/2019 To test the hypothesis is that the mean SAT score for all Stevens High graduates is different from 510 at 5% significance level. a) The null and alternative hypothesis is, Ho: = 510 H,:! #510 The hypothesis test is two-tailed test. b) The t-test statistics is, By using Excel, find t-test statistics with the help of following steps is: 1) Import the data 2) Select the MegaStat from Add-Ins Menu.

Select Mean vs. Hypothesized Value from Hypothesis Tests. Enter Sunmary Input and Hypothesized mean. Select t-test, Alternative and Confidence Level. Click Ok. Hypothesis Test: Mean vs. Hypothesized Value 510.000 hypothesized value 518.975 mean SAT 39.466 std. dev. 6.240 std. error 40 n 39 df 1.44 t 1583 p-value (two-tailed) From the Excel output, the t-test statistics is 1.44). The p-value for this test is,

From the Excel output in part (b), the p-value for this test is 0.1583 d) Decision The conclusion is that the p-value in this context is higher than 0.05 which is 0.1583, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the mean gas mileage is different from 40 miles per gallon. The result is not statistically significant. The answer option is 2).