Question

Suppose 56 out of 100 randomly selected students support the proposal that will
shorten the semester to 15 weeks.
A. What is the sample proportion of supporting the proposal?
B. Under what conditions can we use normal distribution to approximate the
distribution of sample proportion? Verify if these conditions hold.
C. Find a 95% confidence interval for the proportion of all students who support
the proposal.
D. Suppose no students are surveyed. What sample size does the investigator
need to have to achieve a margin of error of at most 3% with 95%
confidence. Suppose we do not have an estimate of population

Answer 1

A)   56 out of 100 randomly selected students support the proposal                          
   Thus sample proportion p̂ = 56/100 = 0.56                          
                              
B)   Normal Approximation can be applied to the distribution of sample proportion under two conditions:                          
   Condition 1: A random sample is taken from a large population.                          
           Since the students are selected randomly, this condition is true                  
    Condition 2: The size of the sample or number of repetitions is relatively large, np and n(1-p) must be at least 10.
           Given n = 100 and p = 0.56, hence np = 56 and n(1-p) = 44. Both are sufficiently large                  
           Hence the condition is satisfied                  
                              
C)   To find 95% confidence interval for true population proportion                          
   For 95%, α = 0.05, α/2 = 0.025                          
   From the z-tables, or Excel function NORM.S.INV(α/2)                          
   z = NORM.S.INV(0.025) = 1.96           (We take the positive value for calculations)              
   Confidence interval is given by                           

   = (0.4627, 0.6573)                          
   95% confidence interval for proportion of all students that support the proposal is (0.4627, 0.6573)                          
                              
D)   For 95% confidence interval, Margin of Error (ME) is given as                          
                              
   ME =    where n is the sample size          
                              
   ME should be atmost 3%                          
   Thus,                          
   
   n > 1051.745                          
   Thus, sample size should be 1052 or more