barcode font reporting services Tests of Hypotheses and Significance in Software

Making QR Code in Software Tests of Hypotheses and Significance

CHAPTER 7 Tests of Hypotheses and Significance
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We take as our decision rule: (1) The claim is not legitimate if Z is less than 2.33 (in which case we reject H0).
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(2) Otherwise, the claim is legitimate, and the observed results are due to chance (in which case we accept H0). !npq !(200)(0.9)(0.1) 4.23. If H0 is true, m np 200(0.9) 180 and s Now 160 in standard units is (160 180) > 4.23 4.73, which is much less than 2.33. Thus by our decision rule we conclude that the claim is not legitimate and that the sample results are highly significant (see end of Problem 7.5). (b) The P value of the test is P(Z 4.73) < 0, which shows that the claim is almost certainly false. That is, if H0 were true, it is almost certain that a random sample of 200 allergy sufferers who used the medicine would include more than 160 people who found relief.
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7.7. The mean lifetime of a sample of 100 fluorescent light bulbs produced by a company is computed to be 1570 hours with a standard deviation of 120 hours. If m is the mean lifetime of all the bulbs produced by the company, test the hypothesis m 1600 hours against the alternative hypothesis m 2 1600 hours, using a level of significance of (a) 0.05 and (b) 0.01. (c) Find the P value of the test.
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We must decide between the two hypotheses H0: m 1600 hours H1: m 2 1600 hours
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A two-tailed test should be used here since m 2 1600 includes values both larger and smaller than 1600. (a) For a two-tailed test at a level of significance of 0.05, we have the following decision rule: (1) Reject H0 if the z score of the sample mean is outside the range (2) Accept H0 (or withhold any decision) otherwise. 1.96 to 1.96.
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# The statistic under consideration is the sample mean X. The sampling distribution of X has a mean mX m and standard deviation sX s> !n, where m and s are the mean and standard deviation of the population of all bulbs produced by the company. Under the hypothesis H0, we have m 1600 and sX s> !n 120> !100 12, using the sample # standard deviation as an estimate of s. Since Z (X 1600)>12 (1570 1600)>12 2.50 lies outside the range 1.96 to 1.96, we reject H0 at a 0.05 level of significance.
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(b) If the level of significance is 0.01, the range 1.96 to 1.96 in the decision rule of part (a) is replaced by 2.58 to 2.58. Then since the z score of 2.50 lies inside this range, we accept H0 (or withhold any decision) at a 0.01 level of significance. (c) The P value of the two-tailed test is P(Z 2.50) P(Z 2.50) 0.0124, which is the probability that a mean lifetime of less than 1570 hours or more than 1630 hours would occur by chance if H0 were true.
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7.8. In Problem 7.7 test the hypothesis m 1600 hours against the alternative hypothesis m a level of significance of (a) 0.05, (b) 0.01. (c) Find the P value of the test.
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We must decide between the two hypotheses H0: m 1600 hours H1: m 1600 hours
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1600 hours, using
A one-tailed test should be used here (see Fig. 7-5). (a) If the level of significance is 0.05, the shaded region of Fig. 7-5 has an area of 0.05, and we find that z1 1.645. We therefore adopt the decision rule: (1) Reject H0 if Z is less than 1.645.
(2) Accept H0 (or withhold any decision) otherwise. Since, as in Problem 7.7(a), the z score is 2.50, which is less than 1.645, we reject H0 at a 0.05 level of significance. Note that this decision is identical with that reached in Problem 7.7(a) using a two-tailed test.
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