A library wants to determine the effectiveness of their summer literacy program

Question

A library wants to determine the effectiveness of their summer literacy program among low-income children. Because surveying the large numbers of students in the program would require too many resources the library staff interviews 30 randomly chosen children among the low-income program attendees. The 30 sampled children are given a reading test before and after the program.

A) The difference in the reading test scores (after – before) has mean 10 and standard deviation 4. Assuming the score differences are normally distributed, what percent of the children showed any improvement (difference > 0) in reading ability?

B) What percent of children improved by more than 15 points?

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    10      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    -2.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -2.5   ) =    0.993790335 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    15      
u = mean =    10      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.25   ) =    0.105649774 [ANSWER]

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