Question

iii. Find the p-value of the observed test-statistic. You need to use the t-table.

t = -1.56

df = n - 1 = 20 - 1 = 19

Answer 1

Hypothesis:

Ho :μ = 15 H, :με 15

Test statistic,

X-portn-1 ta s/n t= 13.5 – 15 4.3/20 t = -1.56

Degrees of freedom = n - 1 = 20 - 1 = 19

P-value = 0.1353 It is obtained using excel with function =TDIST(1.56,19,2)

Now we obtain p- value using t table:

Consider following t table for two tailed hypothesis:

Values of the t-distribution (two-tailed) A 4.785 11 0.80 0.90 0.95 0.980.990.995 0.998 0.999 0.20 0.10 0.05 0.02 0.01 0.005 0.002 0.001 3.078 6.314 12.706 31.820 63.657 127.321 318.309 636.619 1.886 2.920 4.303 6.965 9.925 14.089 22.327 31.599 1.638 2.353 3.182 4.541 5.841 7.453 10.215 12.924 1.533 2.132 2.776 3.747 4.604 5.598 7.173 8 .610 1.476 2.015 2.571 3.365 4.0324.773 5.893 6.869 1.440 1.943 2.447 3.143 3.7074.3175.208 5.959 1.415 1.895 2.365 2.998 3.499 4.029 5.408 1.397 1.860 2.306 2.897 3.355 3.833 4.5011 5.041 1.383 1.833 2.262 2.821 3.250 3.690 4.297 4.781 1.372 1.812 2.228 2.764 3.169 3.581 4.144 4.587 1.363 1.796 2.201 2.718 3.106 3.4974.0254.437 1.356 1.782 2.179 2.681 3.055 3.428 3.930 4.318 1.350 1.771 2.160 2.650 3.012 3.372 3.852 4.221 1.345 1.761 2.145 2.625 2.977 3.326 3.787 4.140 1.341 1.753 2.131 2.602 2.947 3286 3.733 4.073 1.337 1.746 2.120 2.584 2.921 3252 3.686 4.015 1.333 1.740 2.110 2.567 2.898 3.222 3.646 3.965 1.330 1.734 2.101 2.552 2.878 3.1973.610 3.922 1.328 1.729 2.093 2.539 2.861 3.1743.5793.883 1.325 1.725 2.086 2.528 2.845 3.153 3.552 3.850 1.323 1.721 2.080 2.518 2.831 3.135 3.527 3.819 1.321 1.717 2.074 2.508 2.8193.119 3.505 3.792 20 21 22

Now go to degrees of freedom = 19 and find the two number such that calculated absolute value of test statistic lies between them.

Here absolute value of test statistic is 1.56 which lies between 1.328 and 1.729. Now for that corresponding columns see the p value given is metioned in the heading of the column.

Here it is 0.20 and 0.10 Hence

0.10 < p-value < 0.20