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Unformatted text preview: 6. We wish to see if the dial indicating the oven temperature for a certain model oven is properly calibrated. Four ovens of this model are selected at random. The dial on each is set to 300F, and after one hour, the actual tem perature of each is measured. The temperatures measured are 305, 310, 300, and 305. Assume that the distribution of the actual temperatures for this model when the dial is set to 300 is Normal. To test if the dial is properly calibrated, we will test the following hypotheses: H0: = 300 versus Ha: 6 = 300. a. Test the hypotheses at 5% level of significance t test (did it in class) b. Assume the standard deviation for the distribution of actual tempera tures for all ovens of this model when the dial is set to 300 is = 4. At the 5% significance level, what is the power of our test when, in fact, = 310 F ? =P(fail to reject H when H a is true), power=1 First we calculate . Since is know the test statistic is z = x 300 n . So, we fail to reject H if 1 . 96 &lt; x 300 n &lt; 1 . 96. Solve for x . Thus, we fail to reject H if 296 . 08 &lt; x &lt; 303 . 92. = P (296 . 08 &lt; x &lt; 303 . 92 , = 310). Since the population is normally distributes x N ( , n ), beta = P ( 6 . 96 &lt; z &lt; 3 . 04) = P ( z &lt; 3 . 04) P ( z &lt;  6 . 96) = . 0012 0, where the values6.96 and 3.04 are the standardized values for 296.8 and 303.92 respectively. Thus power=1.0012=.9988 c. Find a 95% confidence interval for the true mean. Is the conclusion from a) consistent with the 95% confidence interval? Explain 95% for is (305 3 . 182 * 4 .....
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This note was uploaded on 02/24/2011 for the course STAT 3022 taught by Professor Staff during the Spring '08 term at Minnesota.
 Spring '08
 Staff

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