Question

t nework SetsGrades 1. Which set of hypotheses is appropriate for the foilowing research question is there an significant a and wrilting exam? difference in the average scores of students in the reading 2. Are t the required conditions met to complete this test? Why or wihy not??because The average observed difference in scores is read rie 0.576 and the standard deviation of the differences is 941 points. Do these data prowide convincing evidence of a difference between the average scores on the two exams? Conduct the hypothesis test Round all numeric results to 4 decimal places 3. Calculate the test statistic 4. Calculate the p-value 5.What is your conclusion using α·0.05? A. Reject H B. Do not reject Ho 6.Based on the results of this hypothesis test, would you expect aconfidence iteral for the averageanerence between the indlude 07? 7. Calculate a 95% confidence interval for the average dmerence between the readng and wrong scores of al stoeres adr g and wrtrg scoes to 1418JPG proj 4 reflection docx A MG 1418 (PG ) Type here to search STATE Department of Mathematics Grades ist HW10: Problem 3 Problem Value: 9 point(s). Problem Score: 0% Attempts Remaining: 5 attempts. (9 points) High School and Beyond, Part I. The National Center of Education Statistics conducted a survey of high school seniors, collecting test data on reading, writing, and several other subjects. Here we examine a simple random sample of 200 students from this survey. A histogram of the difference in the reading and writing score of each student is shown below WePBdkS4OEVeAO2user joedanzxetflectiveUser jordanz3 proj 4 reflection docx 1MG-1418.JPG re to search

Answer 1

1

Correct option is A.

2:

Yes data is approximately normally distributed.

(3-5)

Let d = diff = read - write

Here we have d--0.576, 9.41, n 200 Hypotheses are H, Level of significance α 0.05 Test is two tailed Degree of freedom: 1-199 Test Statistics: -(-0.576-0)/(9.41/sqrt(200)) -0.8657 1.972 7 Critical value: If |t| >1.972, Reject HO Rejection Region: Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value-0.3877 P-value: Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis Excel function for critical value Excel function for p-value: TINV(0.05,199) =TDIST(0. 8657,1992)

6:

Since we fail to reject the null hypothesis so confidence interval should contain zero.

7:

Here we have d--0.576, d 9.41 n-200 Since populaiton SD is unknown so t critical value will be used. Level of significance: α=0.05 1.972 Critical value: Degree of freedom: d n-1-199 The requried confidence interval is: dtre =-0.576± (9.41*1.972/sqrt(200)) =-0.576 ± 1.3121 (-1.8881,0.7361) - Excel function for critical value: TINV(0.05,199) Hence, the required confidence interval is (-1.8881,0.7361)