mean bill of $12.99 with a standard deviation of $4.6. Find the 98 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.) The 98% confidence interval is fromA random sample of 140 items is drawn from a population whose standard deviation is known to be ? = 50. The sample mean is formula127.mml = 780. The 98 percent confidence interval is from?Find a confidence interval for ? assuming that each sample is from a normal population. (Round the value of t to 3 decimal places and your final answers to 2 decimal places.)(a) 1formula63.mml = 23, s = 2, n = 7, 90 percent confidence. The 90% confidence interval is (b) 1formula63.mml = 40, s = 4, n = 16, 99 percent confidence. 99 percent confidence.(c) 1formula63.mml = 116, s = 11, n = 26, 95 percent confidence. The 95% confidence interval is Find the margin of error for a poll, assuming that 95% confidence and formula249.mml = 0.3.(a) n = 75 (Round your answer to 4 decimal places.) (b) n = 300 (Round your answer to 4 decimal places.) (c) n = 750 (Round your answer to 4 decimal places.) (d) n = 3,000 (Round your answer to 4 decimal places.)
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