SECTION 6.3 Confidence Intervals for Population Proportions 323 Constructing a Confidence Interval for p The figure below is from a survey of 800 U.S. adults. Construct a 99% confidence interval for the population proportion of U.S. adults who prefer to get their news from television. (Adapted from Pew Research Center) Digital platforms 52% 35% 7% 5% Television Radio Print publications Percent of U.S. adults who prefer to get news from each platform SOLUTION From the figure, np = 0.35. So, nq = 1 - 0.35 = 0.65. Using n = 800, note that nnp = 1800210.352 = 280 7 5 and nnq = 1800210.652 = 520 7 5. So, the sampling distribution of np is approximately normal. Using z c = 2.575, the margin of error is E = zcBnpnq n ≈ 2.575B10.35210.652 800 Use Table 4 in Appendix B to estimate that zc is halfway between 2.57 and 2.58. ≈ 0.043. Next, find the left and right endpoints and form the 99% confidence interval. Left Endpoint Right Endpoint np - E ≈ 0.35 - 0.043 = 0.307 np + E ≈ 0.35 + 0.043 = 0.393 0.307 6 p 6 0.393 x 0.29 0.31 0.33 0.35 0.37 0.39 0.30 0.32 0.34 0.36 0.38 0.40 0.35 0.307 0.393 You can check this answer using technology, as shown at the left. Interpretation With 99% confidence, you can say that the population proportion of U.S. adults who prefer to get their news from television is between 30.7% and 39.3%. TRY IT YOURSELF 3 Use the data in Example 3 to construct a 99% confidence interval for the population proportion of U.S. adults who prefer to get their news from digital platforms. Answer: Page A40 To explore this topic further, see Activity 6.3 on page 329. 6.3 TI-84 PLUS p=0.35 n=800 (.30656,.39344) 1-PropZInt EXAMPLE 3
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