TABLE Values of ta,n 05 025 10 12.706 6.314 3.078 1.886 5.303 2.920 3.182 1.638
ID: 3226713 • Letter: T
Question
TABLE Values of ta,n 05 025 10 12.706 6.314 3.078 1.886 5.303 2.920 3.182 1.638 2.353 2.776 2.132 1.533 2.571 1.476 2.015 2.447 1.895 1.860 1.833 1.363 1.796 2.201 11 1.356 1.782 2.179 12 2.160 1.350 13 2.145 1.761 1.345 14 1.341 1.753 2.131 15 1.746 2.120 1.337 1.740 2.110 1.333 17 1.734 2.101 1.330 18 2.093 1.328 1.729 19 2.086 1.725 1.325 20 2.080 1.721 1.323 21 2.074 1.321 22 2.069 1.714 1.319 2.064 24 1.318 2.060 1.316 1.708 25 2.056 1.706 1.703 2.052 2.048 1.701 1.960 1.645 1.282 Other probabilities: 9864 7635) P176 2.5411 .9825 P178 2.7) 1.66) .94 P(T12 2.8 984 01 31.821 6.965 4.541 3.474 3.365 2.821 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 2.473 2.467 2.462. 2.326 77 PIT11 .934 005 63.657 9.925 5.841 4.604 4.032 3.499 3.355 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 2.576 81 P1711Explanation / Answer
Mean lifetime of certain type of battery = 240 hours
Null Hypothesis : H0 : Mean lifetime of certain type of battery is less than 240 hours. <240.
Alternative Hypothesis : Ha : Mean lifetimes of certain type of battery is atleast 240 hours. >= 240 hours.
Test Statistic :
Mean of the given sample xbar= 237.056 Hours
Standard Deviation of the sample s = 11.28 hours
t = (xbar - 240)/ (s/n) = ( 237.056 - 240)/ ( 11.28/18) = -2.944/2.659 = -1.10
Here alpha = 0.05 and dF = 18-1 = 17 and one tailed test tcritical = t0.05, 17 = 1.740
so t < tcritical, that means we can reject the null hypothesis and we conclude that specification of battery lifetime is at least 240 hours, is met.
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