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Please help me with this

Suppose that a labor union claims that a given procedure requires 20 minutes to perform. To evaluate this, suppose that we measured the amount of time required to complete the procedure on each of 36 occasions.

12, 15, 18, 14, 19, 21, 16, 17, 18, 13, 19, 21, 22, 15, 16, 17, 18, 22, 23, 14, 16, 18, 23, 13, 16, 18, 19, 15, 13, 21, 23, 14, 15, 16, 17, 21

Assuming we wish to be 95% certain of avoiding type 1 error, use these data to evaluate the union’s claim.

hani
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  • How not to do it: in the 36 samples, the time taken is never exactly 20 minutes, so the union is wrong. – Henry Feb 13 '14 at 23:15
  • Can you please elaborate Sire. A proof or something. Thanks – hani Feb 13 '14 at 23:17
  • How to do it: formulate the union's statement into something testable, such as "the average time taken is at least 20 minutes". Then find the probability that you might observe this sample data or something more extreme if the union's claim was correct (you might want to assume a normal distribution). Then see if that probability is less than 5%. – Henry Feb 13 '14 at 23:19
  • You might show us what you have tried – Henry Feb 13 '14 at 23:20
  • Population variance is not known, so normal might not work and we should go for t-distribution. – Satish Ramanathan Feb 13 '14 at 23:26

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Answer:

$H_0: \mu>= 20$

$H_1: \mu< 20$

Test Statistic = (17.4722-20)/(3.167/sqrt(36)= -2.5278/.5278 = -4.78

$t_{{\alpha =0.05},35} = -3.5912$

Since test statistic < t-critical, we reject the null hypothesis (union's claim) and conclude that the procedure requires less than 20 minutes

Satish Ramanathan
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