Question

Boneitis is in the news again: several cases have been reported in Miami, and more are soon to follow. In order to receive aid from the state government, the city of Miami has to provide a small amount of evidence that the prevalence of the disease (i.e., the proportion of Miami residents who have boneitis) is greater than 20%. City ocials took a random sample of 200 residents and found that 33.0% had been diagnosed with boneitis.

(a) (4 points) Perform a z-test at the 0.01 significance level to determine if this sample provides evidence that the prevalence of boneitis in Miami is greater than 20%. Be sure to state the null and alternative hypotheses, any calculator functions you use (including inputs and relevant outputs), and clearly state your conclusion in context.

(b) (3 points) Construct a 95% plus-two confidence interval for pˆ, the sample proportion of Miami residents with boneitis. Show your work, including any calculator functions you use, and be sure to round interval bounds properly.

Answer 1

a)

Ho :   p =    0.2                  
H1 :   p >   0.2       (Right tail test)          
                          
Level of Significance,   α =    0.01                  
Sample Size,   n =    200                  
                          
Sample Proportion ,    p̂ = x/n =    0.3300                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0283                  
Z Test Statistic = ( p̂-p)/SE = (   0.3300   -   0.2   ) /   0.0283   =   4.5962
                          
  
p-Value   =   0.0000   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to conclude that prevalence of boneitis in Miami is greater than 20%

b)α=0.05

Sample Proportion ,    p̂ = x/n =    0.3300          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0332          
margin of error , E = Z*SE =    1.960   *   0.0332   =   0.0652
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.330   -   0.0652   =   0.2648
Interval Upper Limit = p̂ + E =   0.330   +   0.0652   =   0.3952
                  
95%   confidence interval is (   0.265   < p <    0.395   )