Question

An instructor for a test preparation course claims that the course will improve the test scores of students. The table shows the critical reading scores for 14 students the first two times they took the test. Before taking the test for the second time, the students took the instructor's course try to improve their critical reading test scores. At a = 0.01, is there enough evidence to support the instructor's claim? Complete parts (a) through (1) Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Score on first test 427 444 323 534 504 341 368 531 364 390 446 425 628 578 Score on second test 431 446 344 553 461 381 367 545 374 403 460 437 670 604 (a) Identify the claim and state Ho and H, The claim is "The students' critical reading test scores improved the second time they took the test." Let Hd be the hypothesized mean of the students' first score minus their second score. State Ho and H. Choose the correct answer below. O A. Ho Ho sa OB. Hold so C. Ho Hd2d H... H: >0 H... Click to select your answer(s).

Answer 1

c) sd is not equal to 5.437 ,

sd/root(n) =5.437 ( n=sample size =14)  

following are the steps to find sd

first we need the values for d

Sample 1 Sample 2 Difference = Sample 1 - Sample 2 427 431 -4 444 446 -2 323 344 -21 534 553 -19 504 461 43 341 381 -40 368 367 531 545 -14 364 374 -10 390 403 -13 446 460 -14 425 437 -12 628 670 -42 578 604 -26

standard deviaition (d)

N 1 S = (x; – T)? N 1 i=1 $2= 2(X; - x)2 N-1 = (-4--12.357142857143)2 + + (-26 -- 12.357142857143)2 14-1 5379.2142857143 13 = 413.78571428571 S= 1413.78571428571 = 20.341723483661

rest of the test is correct , the test is performed below

From the sample data, it is found that the corresponding sample means are: X = 450.214 X9 = 462.571 Also, the provided sample standard deviations are: 81 = 92.627 89 = 96.546 and the sample size is n = 14. For the score differences we have D = -12.357 8p = 20.342 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Hop 20 Haup < 0 This corresponds to a left-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region Based on the information provided the significance level is a = 0.01, and the degrees of freedom are df = 13. Hence, it is found that the critical value for this left-tailed test is te = -2.65, for a = 0.01 and df = 13. The rejection region for this left-tailed test is R = {t:t< -2.65). (3) Test Statistics The t-statistic is computed as shown in the following formula: t= D sp/vn - 12.357 20.342/V14 -2.273 (4) Decision about the null hypothesis Since it is observed that t = -2.273 > t = -2.65, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.0203, and since p = 0.0203 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean 4 is less than 12, at the 0.01 significance level.

t table for critical value is given below

t.50 t.80 t.95 t.975 t.995 t 9995 cum. prob one-tail two-tails 0.50 1.00 t.75 0.25 0.50 0.20 0.40 t.85 0.15 0.30 t.90 0.10 0.20 0.05 0.10 0.025 0.05 t.99 0.01 0.02 0.005 0.01 t .999 0.001 0.0005 0.002 0.001 df 11 2 3 4 5 6 7 8 9 10 11 121 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 80 100 1000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0% 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.678 0.677 0.675 0.674 50% 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.848 0.846 0.845 0.842 0.842 60% 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.045 1.043 1.042 1.037 1.036 70% 3.078 6.314 12.71 1.886 2.920 4.303 1.638 2.353 3.182 1.533 2.132 2.776 1.476 2.015 2.571 1.440 1.943 2.447 1.415 1.895 2.365 1.397 1.860 2.306 1.383 1.833 2.262 1.372 1.812 2.228 1.363 1.796 2.201 1.356 1.782 2.179 1.350 1.771 2.160 1.345 1.761 2.145 1.341 1.753 2.131 1.337 1.746 2.120 1.333 1.740 2.110 1.330 1.734 2.101 1.328 1.729 2.093 1.325 1.725 2.086 1.323 1.721 2.080 1.321 1.717 2.074 1.319 1.714 2.069 1.318 1.711 2.064 1.316 1.708 2.060 1.315 1.706 2.056 1.314 1.703 2.052 1.313 1.701 2.048 1.311 1.699 2.045 1.310 1.697 2.042 1.303 1.684 2.021 1.296 1.671 2.000 1.292 1.664 1.990 1.290 1.660 1.984 1.282 1.646 1.962 1.282 1.645 1.960 80% 90% 95% Confidence Level 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.390 2.374 2.364 2.330 2.326 98% 63.66 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.660 2.639 2.626 2.581 2.576 99% 318.31 636.62 22.327 31.599 10.215 12.924 7.173 8.610 5.893 6.869 5.208 5.959 4.785 5.408 4.501 5.041 4.297 4.781 4.144 4.587 4.025 4.437 3.930 4.318 3.852 4.221 3.787 4.140 3.733 4.073 3.686 4.015 3.646 3.965 3.610 3.922 3.579 3.883 3.552 3.850 3.527 3.819 3.505 3.792 3.485 3.768 3.467 3.745 3.450 3.725 3.435 3.707 3.421 3.690 3.408 3.674 3.396 3.659 3.385 3.646 3.307 3.551 3.232 3.460 3.195 3.416 3.174 3.390 3.098 3.300 3.090 3.291 99.8% 99.9% Z