Section 9.3 Hypothesis Tests for a Population Mean, Standard Deviation Unknown 435 Step 2: Choose a level of significance. The level of significance is �� = 0.05. Step 3: Compute the test statistic. The test statistic is t = ̄x − ��0 √ n s∕ To compute its value, we note that ̄x = 2.2, s = 6.1, and n = 76. We set ��0 = 0, the value for �� specified by H0. The value of the test statistic is t = 2.2 − 0 √ 76 6.1∕ = 3.144 Step 4: Compute the P-value. When H0 is true, the test statistic t has the Student’s t distribution with n − 1 degrees of freedom. In this case, the sample size is n = 76, so there are n − 1 = 75 degrees of freedom. To obtain the P-value, note that the alternate hypothesis is H1: �� > 0. Therefore, values of the t statistic in the right tail of the Student’s t distribution provide evidence against H0. The P-value is the probability that a value as extreme as or more extreme than the observed value of 3.144 is observed from a t distribution with 75 degrees of freedom. To find the P-value exactly, it is necessary to use technology. The P-value is 0.0012. Figure 9.10 illustrates the P-value as an area under the Student’s t curve, and presents the results from the TI-84 Plus calculator. Step-by-step instructions for performing hypothesis tests with technology are given in the Using Technology section on page 442. Area = 0.0012 3.144 Figure 9.10 The P-value is the area to the right of the observed value of the test statistic, 3.144. The TI-84 Plus display shows that P = 0.0012, rounded to four decimal places. Step 5: Interpret the P-value. The P-value is 0.0012. Because P < 0.05, we reject H0. Step 6: State a conclusion. We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0. Estimating the P-Value from a Table If no technology is available to compute the P-value, the t table (Table A.3) can be used to provide an approximation. When using a t table, we cannot find the P-value exactly. Instead, we can only specify that P is between two values. We now show how to use Table A.3 to bracket P between two values. In Example 9.16, there are 75 degrees of freedom. We consult Table A.3 and find that the number 75 does not appear in the degrees of freedom column. We therefore use the next smallest number, which is 60. Now look across the row for two numbers that bracket the observed value 3.144. These are 2.915 and 3.232. The upper-tail probabilities are 0.0025 for 2.915 and 0.001 for 3.232. The P-value must therefore be between 0.001
navidi_monk_elementary_statistics_2e_ch7-9
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