A common characterization of obese individuals is that their body mass index is

Question

A common characterization of obese individuals is that their body mass index is at least 30 [BMI weight/(height)2, where height is in meters and weight is in kilograms]. An article reported that in a sample of female workers, 264 had BMIs of less than 25, 156 had BMIs that were at least 25 but less than 30, and 122 had BMIs exceeding 30. Is there compelling evidence for concluding that more than 20% of the individuals in the sampled population are obese? (a) State the appropriate hypotheses with a significance level of 0.05 Ho: p = 0.20 Hai p 0.20 Ha: p = 0.20 Ho: p = 0.20 Ha: p > 0.20 O Ho: p 0.20 Hai p +0.20 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = 1.46 p-value = 0.0721 What can you conclude? Do not reject the null hypothesis. There is sufficient evidence that more than 20% of the population of female workers is obese Do not reject the null hypothesis. There is not sufficient evidence that more than 20% of the population of female workers is obese Reject the null hypothesis. There is sufficient evidence that more than 20% of the population of female workers is obese Reject the null hypothesis. There is not sufficient evidence that more than 20% of the population of female workers is obese (b) Explain in the context of this scenario what constitutes type I error A type 1 error would be declaring that 20% or less of the population of female workers is obese, when in fact more than 20% are actually obese A type 1 error would be declaring that 20% or more of the population of female workers is obese, when in fact less than 20% are actually obese A type 1 error would be declaring that less than 20% of the population of female workers is obese, when in fact 20% or more are actually obese A type 1 error would be declaring that more than 20% of the population of female workers is obese, when in fact 20% or less are actually obese

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.24
value of n = 264+ 156 + 122 = 542
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.24*0.76/542)
=0.0183
P( not concludeing more than 20%) = P(X<=20)
P(X > 0.2) = (0.2-0.24)/0.0183
= -0.04/0.0183 = -2.1858
= P ( Z >-2.186) From Standard Normal Table
= 0.9856
P(X < = 0.2) = (1 - P(X > 0.2)
= 1 - 0.9856 = 0.0144

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