Question

t-Distribution Table Find the critical value t, for the confidence level c0 80 and sample size n 9 Click the icon to view the t-distribution table. (Round to the nearest thousandth as needed ) Level of confidence, c 0.80 0.90 0.95 0.98 0.9 One tall a 0.10 0.05 0.025 0.01 0.05 d. Two talls, α 0.20 0.10 005 002 001 du 3.078 6 314 12.706 31.821 63.657 1886 2920 4303 6965 9925 2 638 2.353 3.182 4541 5.841 1.533 2.132 2776 3.747 4604 .476 2.015 2571 3.365 4.032 5 1.440 1.943 2447 3,1433.707 415 1.895 2365 2.998 3.499 1397 1860 2306 2896 3.355 383 1.833 2262 2821 3.2509 1372 1812 2228 2764 3.169 10 363 1796 2201 2.718 3.106 1 1356 1.782 2179 26813.055 12 1350 1.771 2.160 2650 3.012 13 1345 1.761 2145 2.624 2.977 14 341 1.753 2.131 2602 2.947 15 337 1.746 2.120 2583 2.921 333 1,740 2.110 2.567 2.898 17 1.330 1734 2101 2552 2.878 18 1.328 1.729 2.093 2539 2.861 19 1.325 1325 2086 2528 2.845 20 1323 1.721 2080 2518 2831 21 1321 1.717 2074 2508 2819 22 1319 1.714 2.069 2500 2.807 2 1318 1.711 2064 2492 2.797 24 1316 1.708 2.060 2485 2.787 25 314 1.703 2052 2473 2.771 27 313 1.701 2048 2467 2.763 28 311 1.699 2.045 2462 2.756 29 1.310 1.697 2042 2457 2.750 30 1.309 1.696 2.040 2453 2.744 3 1.309 1.694 2037 2449 2.738 32 308 1.692 2.035 2445 2.733 33 Print Done Enter your answer in the answer box

Answer 1

Solution:

Find the critical value t_c for the confidence level c = 0.80 and sample size n = 9.

Answer: The significance level $\alpha=1-c=1-0.80=0.20$ and $df=n-1=9-1=8$ . Therefore, the critical value is:

$\boldsymbol{t_{c}=1.397 }$

Level of confidence, C 90 090 09 098 0.H One tall, LL 0.05 0.025 001 0003 dTwo talls000.10 0.050 001 d 0.01 d.t 3078 6314 12706 ]1.821 63.657 1 1.88612920 4303 6965 9925 2 838 2353 3182 4541 5841 3 1533 2.132 2776 3.747 40 4 476 2015 2571 334 4325 1440 1943 2447 3.145 3.707 1895 2365 2.998 3.4997 397 1.860 306 2896 3.355 8 1.833 2.262 1821 3.250 9 10 1372 1812 2228 2764 3.169 10 1363 1.796 22012718 3.10 13% 1.782 2.179 2681 3.OSS 12 13501.771 2.168 2653 301긴 13 1345 1.761 2145 2624 2.977 14 1341 1.753 2.131 2602 2471 1337 1.746 2120 2583 .921 16 1333 1740 2110 2567 89817 1330 1.734 .101 2552 2878 18 1.꾜8 1.729 2093 Z539 2801 19 1.335 1.72S 2086 2528 2845 20 1323 1.721 2080 2518 2831 21 1.321 1.17 20742508 2819 22 1.319 1.714 2069 2S00 2.807 1.318 1.11 2064 2492 2.797 24 1.316 120 2060 2485 2.787 25 1315 1.796 2056 2479 2.779 26 1314 1.703 2052 473 2.77127 1.313 1.201 2048 2467 2.763 28 1,311 099 204s z462 2.756 29 1310 167 z042 457 2.750 30 309 1.696 Z040 2453 2.744 31 1.309 1094 2037 2449 2.738 32 308 1692 1035 2445 2.733 33 1.307 1691 2032 2441 2.728 34 12 13 141 15 17 18 19 21 23 24 27 281 12

Use the standard normal table to find the z-score that corresponds to the cumulative area 0.0516. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries. Use the z-score halfway between the corresponding z-scores

Answer: Using the table, we have:

$\boldsymbol{z=-1.630 }$