Question
The one-sample t statistic for testing
H0: = 60
Ha: 60
from a sample of n = 29 observations has the value t = 2.68.
(a) What are the degrees of freedom for t?
(b) Locate the two critical values t* from the t distribution critical values table that bracket t.
____< t < ____
(e) If you have software available, find the exact P-value. (Round your answer to four decimal places.)
Table entry for p and C is the critical value t with probability p lying to its right and probability C lying between and t t distribution critical values Upper-tail probability p .25 .20 15 10 05 025 02 01 005 1.000 1.376 1.963 3.078 6.314 12.71 15.89 31.82 63.66 0.816 1.061 1.386 1.886 2.920 4.303 4.849 6.965 9.925 1.638 2.353 3.182 0.765 0.978 1.250 3.482 4.541 5.841 0.741 0.941 1.190 1.533 2.132 2.776 2.999 3.747 4.604 0.727 0.920 1.156 1.476 2.015 2.571 2.757 3.365 4.032 0.718 0.906 1.134 1.440 1.943 2.447 2.612 3.143 3.707 0.896 1.119 1.415 1.895 2.365 2.517 2.998 3.499 0.706 0.889 1.108 1.397 1.860 2.306 2.449 2.896 3.355 0.703 0.883 1.100 1.383 1.833 2.262 2.398 2.821 3.250 0.700 0.879 1.093 1.372 1.812 2.228 2.359 2.764 3.169 11 0.697 0.876 1.088 1.363 1.796 2.201 2.328 2.718 3.106 1.083 1.782 2.179 12 0.695 0.873 1.356 2.303 2.681 3.055 13 0.694 0.870 1.079 1.350 2.160 2.282 2.650 3.012 0.692 0.868 1.076 1.345 1.761 2.145 2.264 2.624 2.977 15 0.691 0.866 1.074 1.341 1.753 2.131 2.249 2.602 2.947 16, 0.690 0.865 1.071 1.337 1.746 2.120 2.235 2.583 2.921 17 0.689 0.863 1.069 1.333 1.740 2.110 2.224 2.567 2.898 18 0.688 0.862 1.067 1.330 1.734 2.101 2.214 2.552 2.878 0.688 0.861 1.066 1.328 1.729 2.093 2.205 2.539 2.861 20 0.687 0.860 1.064 1.325 1.725 2.086 2.197 2.528 2.845 21 0.686 0.859 1.063 1.323 1.721 2.080 2.189 2.518 2.831 1.717 2.074 22 0.686 0.858 1.061 1.321 2.183 2.508 2.819 23 0.685 0.858 1.060 1.319 1.714 2.069 2.500 2.807 24 0.685 0.857 1.059 1.318 2.064 2.172 2.492 2.797 1.708 2.060 25 0.684 0.856 1.058 1.316 2.167 2.485 2.787 26 0.684 0.856 1.058 1.315 1.706 2.056 2.162 2.479 2.779 27 0.684 0.855 1.057 1.314 1.703 2.052 2.158 2.473 28 0.683 0.855 1.056 1.313 1.701 2.048 2.154 2.467 2.763 29 0.683 0.854 1.055 1.311 1.699 2.045 2.150 2.462 2.756 30 0.683 0.854 1.055 1.310 1.697 2.042 2.147 2.750 40 0.681 0.851 1.050 1.303 1.684 2.021 2.123 2.423 2.704 50 0.679 0.849 1.047 1.299 1.676 2.009 2.109 2.403 2.678 0.679 0.848 1.045 1.296 1.671 2.000 2.099 2.390 2.660 80 0.678 0.846 1.043 1.292 1.664 1.990 2.088 2.374 2.639 100 0.677 0.845 1.042 1.290 1.660 1.984 2.081 2.364 2.626 1000 0.675 0.842 1.037 1.282 1.646 1.962 2.056 2.330 2.581 0.674 0.841 1.036 1.282 1.645 1.960 2.054 2.326 2.576 50% 60% 70% 80% 90% 95% 96% 98% 99% Confidence level C Probability p 0025 001 127.3 318.3 14.09 22.33 7.453 10.21 5.598 7.173 4.773 5.893 4.317 5.208 4.029 4.785 3.833 4.501 3.690 4.297 3.581 4.144 3.497 4.025 3.428 3.930 3.372 3.852 3.326 3.787 3.286 3.733 3.252 3.686 3.222 3.646 3.197 3.611 3.174 3.579 3.153 3.552 3.135 3.527 3.119 3.505 3.104 3.485 3.091 3.467 3.078 3.450 3.067 3.435 3.057 3.421 3.408 3.047 3.038 3.396 3.030 3.385 2.971 3.307 2.937 3.261 2.915 3.232 2.887 3.195 2.871 3.174 2.813 3.098 2.807 3.091 99.5% 99.8% 0005 636.6 31.60 12.92 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 3.819 3.792 3.768 3.745 3.725 3.707 3.690 3.674 3.659 3.646 3.551 3.496 3.460 3.416 3.390 3.300 3.291 99.9%
Explanation / Answer
Given ,
t=2.68
a) here degree of freedom is n-1=21-1=20
b) this two tailed test because alternative hypothesis contain not equal to sign.
We write in .
-2.68< t< 2.68
c)
From table see in first colomn is df . choose df=20
And that row find out two value lie 2.68 that values .
In table you 2.68 lie between 2.528 and 2.845
So pvalue lie between 0.005 and 0.01
For using excel or ti-83/84 you get exact p value
P value =0.0144
