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Unformatted text preview: &¡¢£¢ ¤¥ &¦§¨©ª¨«¬­®«¯­¦°± ¡­²³©ª­´µ 8.1 In this exercise, the parameter of interest is μ , the population mean. The objective of the experiment is to show that the mean exceed 2.3. a. We want to prove the alternative hypothesis that μ is, in fact, greater than 2.3. Hence, the alternative hypothesis is H a : μ > 2.3 and the null hypothesis is H : μ = 2.3 b. the probability of a Type I error is defined as & = P[reject H 0 when H is true] = P[decide μ > 2.3 when μ = 2.3] c. The best estimator for μ is the sample average x , and the test statistics is n x z σ μ- = which represent the distance (measured in units if standard deviations) from x to the hypothesized mean μ . In order to perform the test at significance level & =.05, given the hypotheses above, we complete the following steps: (i) The value of the test statistic is 04 . 2 35 / 29 . 3 . 2 4 . 2 * =- = z (ii) P-value = ( ) ( ) 0207 . 04 . 2 * = > = > Z P z Z P , which we find with the help of the Normal Table (iii) The p-value of 0.0207 is smaller than the significance level ( & =.05). Therefore, we reject the null hypothesis (iv) The conclusion is that there is evidence in the sample that the population mean μ is larger than 2.3. 8.2. If the experimenter wishes to prove that μ < 2.9, this is the alternative hypothesis. Hence, the hypothesis to be tested is H : μ = 2.9 H a : μ < 2.9 and the test is one-tailed. Let’s perform the test at significance level & =.05. (i) The value of the test statistic (the same z-statistic from 8.1), is 2 . 10 35 / 29 . 9 . 2 4 . 2 *- =- = z (ii) P-value = ( ) 2 . 10 ≈- < Z P (iii) The p-value of 0 is smaller than the significance level level ( & =.05). Therefore, we reject the null hypothesis (iv) The conclusion is that there is evidence in the sample that the population mean μ is smaller than 2.9. 8.3. If the experimenter wishes to prove that μ & 2.9, this is the alternative hypothesis. Hence, the hypothesis to be tested is H : μ = 2.9 H a : μ & 2.9 and the test is two-tailed....
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