Question: Answer try it yourself questions 1-4 TI-83/84 1 on EXAMPLE 2 361. Constructing a Confidence Interval for p Construct a 95% confidence interval for the

Answer try it yourself questions 1-4

Answer try it yourself questions 1-4 TI-83/84 1 on EXAMPLE 2 361.Constructing a Confidence Interval for p Construct a 95% confidence interval forthe proportion of adults in the United States who say that theirfavorite sport to watch is football. Solution From Example 1, p ~

TI-83/84 1 on EXAMPLE 2 361. Constructing a Confidence Interval for p Construct a 95% confidence interval for the proportion of adults in the United States who say that their favorite sport to watch is football. Solution From Example 1, p ~ 0.290402. So, q = 1 - 0.290402 = 0.709598. Using n = 1219, you can verify that the sampling distribution of p can be approximated by the normal distribution. np ~ 1219 . 0.290402 ~ 354 > 5 and np ~ 1219 . 0.709598 ~ 865 > 5 Using Ze = 1.96, the margin of error is pq ~ 1.961 (0.290402) (0.709598) E = Zc Vn ~ 0.025. 1219 The 95% confidence interval is as follows. ip le 2 that Left Endpoint Right Endpoint interval P - E = 0.29 - 0.025 = 0.265 p + E = 0.29 + 0.025 = 0.315 on p is e decimal 0.265

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