8-2 Testing a Claim About a Proportion 393 3. The requirements np Ú 5 and nq Ú 5 are both satisfied with n = 926, p = 0.5, and q = 0.5. [The value of p = 0.5 comes from the claim. We get np = 1926210.52 = 463 and we get nq = 1926210.52 = 463, so both np and nq are greater than or equal to 5.] The three requirements are satisfied. Solution: P-Value Method Technology: Computer programs and calculators usually provide a P-value, so the P-value method is commonly used. See the accompanying TI-83>84 Plus calculator results showing the alternative hypothesis of “prop 7 0.5,” the test statistic of z = 1.25 (rounded), and the P-value of 0.1059 (rounded). Table A-2: If technology is not available, Figure 8-1 on page 376 in the preceding section lists the steps for using the P-value method. Using those steps from Figure 8-1, we can test the claim that most Internet users utilize two-factor authentication to protect their online data. We use the survey results of n = 926 and pn = 0.52. Step 1: The original claim is that most Internet users utilize two-factor authentication to protect their online data. That claim can be expressed in symbolic form as p 7 0.5. Step 2: The opposite of the original claim is p … 0.5. Step 3: Of the preceding two symbolic expressions, the expression p 7 0.5 does not contain equality, so it becomes the alternative hypothesis. The null hypothesis is the statement that p equals the fixed value of 0.5. We can therefore express H0 and H1 as follows: H0: p = 0.5 H1: p 7 0.5 (original claim) Step 4: For the significance level, we select a = 0.05, which is a very common choice. Step 5: Because we are testing a claim about a population proportion p, the sample statistic pn is relevant to this test. The sampling distribution of sample proportions pn can be approximated by a normal distribution in this case. Step 6: The test statistic z = 1.25 can be found by using technology or it can be calculated by using pn = 482>926, n = 926 (sample size), p = 0.5 (assumed in the null hypothesis), and q = 1 - 0.5 = 0.5. (482 is 52% of 926; see the hint on page 391.) z = pn - p Apq n = 482 926 - 0.5 B10.5210.52 926 = 1.25 The P-value can be found from technology or it can be found by using the following procedure, which is shown in Figure 8-3 on page 380. Left-tailed test: P-value = area to left of test statistic z Right-tailed test: P-value = area to right of test statistic z Two-tailed test: P-value = twice the area of the extreme region bounded by the test statistic z Because this hypothesis test is right-tailed with a test statistic of z = 1.25, the P-value is the area to the right of z = 1.25. Referring to Table A-2, we see that the cumulative area CP TI-83, 84 Plus Is 0.05 a Bad Choice? The value of 0.05 is a very common choice for serving as the cutoff separating results considered to be significant from those that are not. Science writer John Timmer wrote in Ars Technica that some problems with conclusions in science are attributable to the fact that statistics is sometimes weak because of the common use of 0.05 for a significance level. He gives examples of particle physics and genetics experiments in which P-values must be much lower than 0.05. He cites a study by statistician Valen Johnson, who suggested that we should raise standards by requiring that experiments use a P-value of 0.005 or lower. We do know that the choice of 0.05 is largely arbitrary, and lowering the significance level will result in fewer conclusions of significance, along with fewer wrong conclusions.
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