According to a survey in a country 32 of adults do not have

According to a survey in a country, 32% of adults do not have any credit cards. Suppose a simple random sample of 800 adults is obtained.

Determine the mean of the sampling distribution of (phat) p ____

Determine the standard deviation of the sampling distribution of (phat) p_____

In a random sample of 800 adults, what is the probability that less than 30% have no credit cards? The probability is ______

Solution

Determine the mean of the sampling distribution of (phat) p ____

As stated already, p^ = 0.32 [ANSWER]

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Determine the standard deviation of the sampling distribution of (phat) p_____

As

s(p) = sqrt[p(1-p)/n)]

Then

s(p) = sqrt(0.32*(1-0.32)/800) = 0.016492423 [ANSWER]

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In a random sample of 800 adults, what is the probability that less than 30% have no credit cards?

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
po = critical value =    0.3      
p = mean =    0.32      
          
s(p) = standard deviation =    0.016492423      
          
Thus,          
          
z = (po - p) / s(p) =    -1.212678089      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -1.212678089   ) =    0.11262646 [ANSWER]