Question

(1 point) Use the following information for all parts of this problem. On a particular golf course, a sample of 44 golfers have a mean golf score of 83. Suppose the population standard deviation for this course is 4.72. Part 1: Part 2: Using the formula from part 1, find a 95% confidence interval for the mean golf score of all golfers. Round each answer to the nearest hundredth (2 decimal places). 81.61 to 84.39 Part 3: Calculate the margin of error for the answer in part 2. Round the answer to the nearest hundredth (2 decimal places).

Answer 1

use 1 one the following information for all parts of this problem. a particular golf course a sample of 44 golfers have a mean golf score of 83. Suppose the population Standard deviation for this course is 4.72 Here are given, Sample size, n = 44 Sample mean, 5 - 83 population standard deviation, o = =4.72 Part 2 using the formula, find a 95% confidence interval for the mean golf score of all golfers. Round each answer to nearest hundredth 2 decimal places) the 100 (1-2) confidence interval for given population by. mean (l) is 5c - Zolia Tala o Luc LULJC + Zdja 6145 Here x=0.05, = 0.025 Za 20.025 11 1.96 [From Standard a Normal distribution table ] 0.025 0.025 0.95 mhain z = -1.9C -0.025 20.025 -1.96

2 Therefore the 95% confidence interval for golfers (M) is mean golf score of all given by OC - 1.96 <Mañ+ 1.96 - 1 V 83 - 1.96 x 4.72 Luz 83+ 1.96%. 544 4.72 544 > 83 - 9.9512 544 L u <83+ 9.2512 144 → 83 - 1.39467 L M <83+ 1.39467 81.60533 L M L 84.39467 V 81.61 L M L 84.39 (Round off 2 decimal places) > UE [81.61, 84.39] interval for for mean the 95% confidence golfers is given by golf score of all 81.61 to 84.39 Ang PART-3 margin of error for the answer in part 2. Round the answer to the nearest hundredth (2 decimal places) calculate the Margin of error 142 Sn 1.96 x 4.72 144 9.2512 544 1.39467 1.39 (Round off a decimal places) Therefore, the margin of error for the answer in part 2 is given by 1.39 Ans