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One and Two-Sample Hypothesis Testing

A manufacturer has developed a new fishing line, which he claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilogram. To test the hypothesis that &#956; = 15 kilograms against the alternative that &#956; < 15 kilograms, a random sample of 50 lines will be tested. The critical region is defined to be x < 14.9.

a) Find the probability of committing a type I error when H0 is true.
b) Evaluate &#946; for the alternatives &#956; = 14.8 and &#956; = 14.9 kilograms.

18. If we plot the probabilities of failing to reject H0 corresponding to various alternatives for &#956; (including the value specified by H0) and connect all the points by a smooth curve, we obtain the operating characteristic curve of the test criterion, or simply the OC curve. Note that the probability of failing to reject H0 when it is true is simply 1 - &#945;. Operating characteristic curves are widely used in industrial applications to provide a visual display of the merits of the test criterion. With reference to Exercise 15, find the probabilities of failing to reject H0 for the following 9 values of &#956; and plot the OC curve: 184, 188, 192, 196, 200, 204, 208, 212, and 216.

Exercise 15: A soft-drink machine at a steak house is regulated so that the amount of drink dispensed is approximately normally distributed with a mean of 200 milliliters and a standard deviation of 15 milliliters. The machine is checked periodically by taking a sample of 9 drinks and computing the average content. If x falls in the interval 191 < x < 209, the machine is thought to be operating satisfactorily; otherwise, we conclude that &#956; does not equal 200 milliliters.

a) Find the probability of committing at type I error when &#956; = 200 milliliters.
&#956; = 200, n = 9, &#417; = 15 and &#417; x = 15/3 = 5
So, z1 = (191-200)/5 = -1.8, and z2 = (209 - 200)/5 = 1.8,
with a &#945; = 2P(Z < -1.8) = (2)(0.0359) = 0.0718

b) Find the probability of committing a type II error when &#956; = 215 milliliters.
If &#956; = 215, then z - 1 = (191-215)/5 = -4.8 and z2 = (209-215)/5 = -1.2, with
&#946;= P(-4.8 < Z < -1.2) = 0.1151 - 0 = 0.1151

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One- and Two-Sample Hypothesis Testing are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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