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, 157 had BMIs that were at least 25 but less than 30, and 121 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.

H₀: p = 0.20
H_a: p ≠ 0.20H₀: p > 0.20
H_a: p = 0.20    H₀: p = 0.20
H_a: p < 0.20H₀: p = 0.20
H_a: 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 = _______.

P-Value = _______.

What can you conclude?

Reject the null hypothesis. There is not sufficient evidence that more than 20% of the population of female workers is obese.Do not 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 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.

(b) Explain in the context of this scenario what constitutes type I error.

A type I 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 I 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 I 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 I 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.

Explain in the context of this scenario what constitutes type II error.

A type II 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 II 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 II 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 II 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.

(c) What is the probability of not concluding that more than 20% of the population is obese when the actual percentage of obese individuals is 26%? (Round your answer to four decimal places.)

Answer 1

a)

H₀: p = 0.20
H_a: p > 0.20

Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   121                  
Sample Size,   n =    264                  
                          
Sample Proportion ,    p̂ = x/n =    0.4583                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0246                  
Z Test Statistic = ( p̂-p)/SE = (   0.4583   -   0.2   ) /   0.0246   =   10.49
                          
  
p-Value   =   0.0000   [Excel function =NORMSDIST(-z)              

Decision:   p-value<α , reject null hypothesis   

Reject the null hypothesis. There is sufficient evidence that more than 20% of the population of female workers is obese

b)type I error -

A type I 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.

-------------------

type II error-

A type II 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

c)

true proportion,   p=   0.26                      
                              
hypothesis proportion,   po=    0.2                      
significance level,   α =    0.05                      
sample size,   n =   264                      
                              
std error of sampling distribution,   σpo = √(po*(1-po)/n) = √ (   0.200   *   0.800   /   264   ) =   0.0246
std error of true proportion,   σp = √(p(1-p)/n) = √ (   0.26   *   0.74   /   264   ) =   0.0270

Zα =       1.645   (right tailed test)      
                  
We will fail to reject the null (commit a Type II error) if we get a Z statistic <                   1.645
this Z-critical value corresponds to X critical value( X critical), such that                  
                  
(p^ - po)/σpo ≤ Zα                  
p^ ≤ Zα*σpo + po                  
p^ ≤    1.645*0.0246+0.2       =   0.2405  
                  
now, type II error is ,ß =    P( p^ ≤    0.2405   given that p =   0.26  
                  
   = P ( Z < (p^ - p)/σp )=       P(Z < (0.2405-0.26) / 0.027)      
   = P ( Z < (   -0.723   )      
ß   =   0.2350 (answer)