Question

SAT For a specific year, the national average score on the SAT Math test was 515. 496 A superintendent wishes to see if their students scored, on average, significantly below the national average on the test. They randomly selected 36 student scores, as shown. 505 522 519 287 243 506 (a) (3 points) At a 0.10 significance level, perform at test to test if there is enough evidence to support the superintendent's claim. Provide the output below starting at the Yellow highlighted cell. 522 668 516 523 509 507 531 543 714 576 523 505 762 519 517 516 503 438 513 349 511 515 414 499 (b) (3 points) Compute a 90% confidence interval for the average SAT score of the students of this superindent. Provide your final anwers in the yellow highlighted cells below. ( Keep 3 or more decimal places) Show your work below those cells. 493 506

Please help me with what formulas to put in to get the right answers. I have tried but I am stuck not getting it... For the first one I am only able to find a two paired t-test but there is only one row so I don't know what to do...

Answer 1

I) a) a t-test allows us to compare the average values of the two data sets and determine if they came from the same population.

using t test here we wat to test mu=515 against mu=! 515

our test statistics is

X – μ t= Z — S ô / vn

b)confidence interval is Lower limit = M - Z.95σM

Upper limit = M + Z.95σM

where M is the mean

II) a)
The two-sample t-test for unpaired data is defined as: Ho: Mi = 42 Mi + u2 Test Statistic: Y-Y, s/Ni+s/N, Hai T= where Ny and Ny are the sample sizes, 71 and Y2 are the sample means, and sĩ and sź are the sample variances. If equal variances are assumed, then the formula reduces to: T = Y-Y, *P/1/+1/NĐ where (N-1)s; +(N-1)s; Ni+N, -2 Significance a. Level: Critical Reject the null hypothesis that the two means are equal if Region: |T1 > 11-a/2,v

b)