Please show all work. Thank you 1. Use KS test to test the following hypotheses

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Please show all work. Thank you
1. Use KS test to test the following hypotheses at = 0.05; HO: Failure times are normally distributed with mean 10 and standard deviation of 5; H: Failure times are NOT normally distributed with mean 10 and standard deviation of all 1sisn We reject the null hypothesis, when DMAXDERITICAL Tablel. Calculation of KS Test TTF F(t) F(t)-(i-1)/n /n-F(MAX 0.2 0.0250 0.0250 0.9 0.0344 -0.0056 .8 0.1075 0.0275 4 0.3015 0.1815 7.6 0.3156 0.1556 8.2 0.3594 0.1594 8.6 0.3897 0.1497 9.1 0.4286 0.1486 .3 0.4443 0.1243 9.4 0.4522 0.0922 9.5 0.4602 0.0602 10.3 0.5239 0.0839 10.9 0.5714 0.0914 0,0150 0.025 0,0456 0.0456 0.0125 0.027 0.1415 0.1815 0.156 0.1556 0.1194 0.1594 0.1097 0.1497 0.1086 0.1486 0.0843 0.1243 0.0522 0.0922 0,0202 0.060 0.0439 0.0839 0,0514 0.0914 0.0655 0.1055 0.0255 0.0655 0.0080 0.0480 0.0100 0.0300 0.0057 0.0457 0.0020 0.0380 .0236 0.0236 10 13 14 11.6 0.6255 0.1055 15 11.6 0.6255 0.0655 11.9 0.6480 0,0480 1122 0.6700 0.0300 18 19 20 21 3 0.7257 0.0457 13.5 0.7580 0.0380 13.8 0.7764 0.0164 14 0.7881 -0.0119 0,0519 0,0519 4.1 0.7939 -0,0461 .086 0.0861 23 14.4 0.8106 -0,0694 0.1094 0.1094 24 25 22.6 0.994 0.034 6.8 0.913 -0,0069 0.0469 0.0469 0,0059 0,0341

Explanation / Answer

We reject when Dmax > Dcritical

The max value of Dmax = .1815

checking p-value of the test from the table for level = .05

One-sample Kolmogorov-Smirnov test

p-value = 0.4431

alternative hypothesis: two-sided

As the p-value of the test is >.05, we shall not reject the null hypothesis.

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