Question

In: Statistics and Probability

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7. Part C....

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.

Part C. Five of the 20 lakes have a pH less than 4.5.
Part Ci. Build a 95% confidence interval for the proportion of lakes that have a pH less than 4.5.
Part Cii. Test whether the population proportion of pH’s that are less than 4.5 differs from 0.5.
Part Ciii. To be accurate to within 0.1 with 95% confidence, how many lakes should we measure?

Solutions

Expert Solution

ci)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   5          
Sample Size,   n =    20          
                  
Sample Proportion ,    p̂ = x/n =    0.2500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0968          
margin of error , E = Z*SE =    1.960   *   0.0968   =   0.1898
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.250   -   0.1898   =   0.060
Interval Upper Limit = p̂ + E =   0.250   +   0.1898   =   0.440
                  
95%   confidence interval is (   0.0602   < p <    0.440   )
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cii)

Ho :   p =    0.5                  
H1 :   p ╪   0.5       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   5                  
Sample Size,   n =    20                  
                          
Sample Proportion ,    p̂ = x/n =    0.2500                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.1118                  
Z Test Statistic = ( p̂-p)/SE = (   0.2500   -   0.5   ) /   0.1118   =   -2.2361
                          
  
p-Value   =   0.0253   [excel formula =2*NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to conclude that   population proportion of pH’s that are less than 4.5 differs from 0.5   

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ciii)

sample proportion ,   p̂ =    0.25                          
sampling error ,    E =   0.1                          
Confidence Level ,   CL=   0.95                          
                                  
alpha =   1-CL =   0.05                          
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.1   ) ² *   0.25   * ( 1 -   0.25   ) =    72.03
                                  
                                  
so,Sample Size required=       73                          


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