Question

In: Statistics and Probability

A. You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p∗=28%p∗=28%.

 

A. You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p∗=28%p∗=28%. You would like to be 99.5% confident that your esimate is within 2.5% of the true population proportion. How large of a sample size is required?n =___

B. Giving a test to a group of students, the grades and gender are summarized below

  A B C Total
Male 17 15 16 48
Female 7 6 9 22
Total 24 21 25 70

Let ππ represent the percentage of all female students who would receive a grade of B on this test. Use a 95% confidence interval to estimate ππ to three decimal places.
Enter your answer as a tri-linear inequality using decimals (not percents). ___

C. Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 111 with 49 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 98% C.I. =___

Solutions

Expert Solution

A)

sample proportion ,   p̂ =    0.28                          
sampling error ,    E =   0.025                          
Confidence Level ,   CL=   0.995                          
                                  
alpha =   1-CL =   0.005                          
Z value =    Zα/2 =    2.807   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   2.807   /   0.025   ) ² *   0.28   * ( 1 -   0.28   ) =    2541.59
                                  
                                  
so,Sample Size required=       2542                          

B)

Level of Significance,   α =    0.05  
Number of Items of Interest,   x =   6  
Sample Size,   n =    70  
          
Sample Proportion ,    p̂ = x/n =    0.086  
z -value =   "Zα/2 =
"   1.9600   [excel formula =NORMSINV(α/2)]
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0335  
          
margin of error ,    E = Z*SE =    0.0656

  
          
Confidence Interval          
Interval Lower Limit , =    p̂ - E =    0.0201  
Interval Upper Limit , =    p̂ + E =   0.1513  

(0.020

c)

Level of Significance,   α =    0.02  
Number of Items of Interest,   x =   49  
Sample Size,   n =    111  
          
Sample Proportion ,    p̂ = x/n =    0.441  
z -value =   "Zα/2 =
"   2.3263   [excel formula =NORMSINV(α/2)]


Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0471  
          
margin of error ,    E = Z*SE =    0.1096  
          
Confidence Interval          
Interval Lower Limit , =    p̂ - E =    0.3318  
Interval Upper Limit , =    p̂ + E =   0.5511  

(0.332


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