Question

A common characterization of obese individuals is that their body mass index is at least 30 (BMI = weight/(height)?, where height is in meters and weight is in kilograms]. An article reported that in a sample of female workers, 266 had BMIs of less than 25, 158 had BMIs that were at least 25 but less than 30, and 123 had BMIs exceeding 30. Is there compelling evidence for concluding that more than 20% of the individuals in the sampled population are obese?

Answer 1

Solution Given that An article reported that in a sample of female workers, 266 had Bm Is of less than 25, 158 had Bm Is that were atleast 25 but - less than 30. 123 had BMTs exceeding 30- The observed proportion is, ê= 123 266+158+123 123 547 TÊ = 0.2249 Is there compelling evidence for concluding that more than 20% of individuals in the sampled population are obse? Normal approximation can be used when mpo 210 and nqo210. from given in dormation, n=547, P-0-2,.q=0.8 where ne... 547X D = 109.4 >10

nao=547X0.8 . = 437.6 210 Novo, we can use normal approximation. The value of test statistic: P-Po NPo (1-Po)/n - [P=0.2] - 0.2249-0.2 N0.21-0.9)/547 0.0249 0.017103 TZ = 1.45597 ♡ 1.46, Take 2=0.05 then, The critical value is ) Zorce11645 (-from slanband o normal table) ... the critical region is 2 71,645 Since 1.46 <1:645, do not reject ruli hypothesis. → There is not enough evidence to conclude that the proportion of obese people exceeds 20%.

Here, There are 2 types of error Type I error: The proportion is more than 2014 when this is not the case. Type I error: The proportion qs 2018when the actual proportion is more than 20% (C) The probability of not concluding that more than 2014. of the population is obese when the actual percentage of obese individuals is 27.4. Take, The probability of a type Tedros i8 , $(0.27): 0 Po-P'+ Zen Poll-Polin ✓ P" (1-P') In 20/0.2 - 0.27 +1645,0-221-0. 2)/547 NO.271-0. 27547 ] *°/-0.0418667 a 0.01898)

= $(2.9058) = 0.0137 1 - from Normal distribution 1. table) Therefore, The probability of a type II error is 0.0137