1) A grower believes that one in five of his citrus trees are infected with the

Question

1) A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within 0.07 with probability 0.95?

2) A random sample of n = 400 observations from a binomial population produced x = 161 successes. Find a 90% confidence interval for p.

from....to....

3) A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. They expect that about 75% of the student body would respond favorably.

(a) What sample size is required to obtain a 95% confidence interval with an approximate margin of error of 0.046?

(b) Suppose that 55% of the sample responds favorably. Calculate the margin of error for the 95% confidence interval.

Explanation / Answer

1.

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.025  
       
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
E =    0.07  
p =    0.2  
      
Thus,      
      
n =    125.4353901  
      
Rounding up,      
      
n =    126   [ANSWER]

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2.

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.4025          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.024520081          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.040331944          
lower bound = p^ - z(alpha/2) * sp =   0.362168056          
upper bound = p^ + z(alpha/2) * sp =    0.442831944          
              
Thus, the confidence interval is              
              
(   0.362168056   ,   0.442831944   ) [ANSWER]

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