R Assignment 7

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School

Montgomery College *

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100

Subject

Statistics

Date

Feb 20, 2024

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docx

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Uploaded by GeneralElk1670

Kevin Pham STAT100 Section 0331 R Assignment 7 1.a. 1.b. 26.962 ± 1.96 (7.143/√92) = 26.962 ± 1.96 (0.7447) = 26.962 ± 1.460 = (25.502, 28.422) 1.c. = (25.502 + 28.422)/2 𝑝̂ = 26.962 1.d. E = (28.422 - 25.502)/2 = 1.46 1.e. We are 95% confident that the population mean body fat percentage of adolescent girls is between 25.502 and 28.422 percent. 1.f. 26.962 ± 2.576 (7.143/√92) = 26.962 ± 2.576 (0.7447) = 26.962 ± 1.918 = (25.044, 28.880) 1.g.

= (25.044 + 28.880)/2 𝑝̂ = 26.962 1.h. (28.880 - 25.044)/2 = 1.918 2.a. 2.b. 0.826 ± 1.645 (√ (0.826) (1-0.826)/92) = 0.826 ± 1.645 (√ 0.14372/92) = 0.826 ± 1.645 (√ 0.001562) = 0.826 ± 1.645 (0.040) = 0.826 ± 0.0658 = (0.7602, 0.8918) 2.c. = (0.8918 + 0.7602)/2 𝑝̂ = 0.826 2.d. E = (0.8918 - 0.7602)/2 = 0.066 2.e. We are 90% confident that the population proportion of adolescent girls with a “medium” activity level is between 0.7602 and 0.8918. 3.a. (6.85 + 7.49)/2

= 7.17 Since we want to find the sample mean, we take the confidence interval and add them. Then we divide the answer by two. 3.b. 6.96 + 0.21 = 7.17 Since the margin of error is the difference between the two endpoints, you add the margin of error to the lower endpoint to get the upper endpoint. 3.c. N = 1/(0.0225)^2 = 1976

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R Assignment 7

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