A golf ebb currently has 425 members that pay annual dues to play golf on the co

Question

A golf ebb currently has 425 members that pay annual dues to play golf on the course. Historical records show that 36% of the members use the course at least once a week during the summer season. To verify this, a random sample of 100 members was selected. Complete parts a through d below. a. Calculate the standard error of the proportion. sigma_p = b. What is the probability that 29 or more members from the sample used the course at least once a week? P(29 or more members) = c. What is the probability that between 30 and 40 members from the sample used the course at least once a week? P(Between 30 and 40 members) = d. If 48 members fiom the sample used the course at least once a week, does this support the historical records of course usage? Select the correct choice below and, fill in the answer box to complete your choice. A. No, since the probability of obtaining a result this small or smaller is, given that the historical records give an accurate estimation of course usage. B. Yes, since the prbability of obtaining a result this large or larger is, given that the historical records give an accurate estimation of

Explanation / Answer

a)

Here,          
n =    100      
p =    0.36      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.048      
As for the finite population correction, N = 425,

fpc = sqrt[(N-n)/(N-1)] = 0.875505245

hence, the standard error is

s(p^) = 0.048*0.875505245 = 0.042024252 [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 = 29/100 =   0.29      
u = mean = p =    0.36      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.042024252      
          
Thus,          
          
z = (x - u) / s =    -1.665704841      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.665704841   ) =    0.952113891 [ANSWER]

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c)

Here,          
n =    100      
p =    0.36      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.3      
x2 = upper bound =    0.4      
u = mean = p =    0.36      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.042024252      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.427747007      
z2 = upper z score = (x2 - u) / s =    0.951831338      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.076682345      
P(z < z2) =    0.829408737      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.752726391   [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.48      
u = mean = p =    0.36      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.042024252      
          
Thus,          
          
z = (x - u) / s =    2.855494013      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.855494013   ) =    0.002148497 [ANSWER]

Hence,

NO, since the probability of obtaining a result this large or larger is 0.002148497, given that the historical records give an accurate estimation of the course usage. [CONCLUSION]
  

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Hi! Note that I did not use continuity correction here. If you use that in class, please resubmit this question stating to use continuity correction. That way we can continue helping you! Thanks!

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