According to a survey in a country, 38% of adults do not have any credit cards.

Question

According to a survey in a country, 38% of adults do not have any credit cards. Suppose a simple random sample of 800 adults is obtained.

1. Determine the mean of the sampling distribution of "p hat"

2. Determine the standard deviation of the sampling distribution of "p hat"

3. In a random sample of 800 adults, what is the probability that less than 36% have no credit cards?

-Would it be unusual if a random sample of 800 adults results in 320 or more having no credit cards?

a. The result is not usual because the probability that "p hat" is greater than or equal to this sample proportion is less than 5%

b. The result is not usual because the probability that "p hat" is greater than or equal to this sample proportion is greater than 5%

c. The result is usual because the probability that "p hat" is greater or equal to this sample proportion is greater than 5%

d. The result is usual because the probability that "p hat" is greater or equal to this sample proportion is less than 5%

Explanation / Answer

1.

The mean is p itself,

u(p^) = 0.38 [ANSWER]

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2.

Also,

s(p^) = sqrt(p(1-p)/n) = sqrt(0.38*(1-0.38)/800) = 0.017161002 [ANSWER]

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3.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.36      
u = mean =    0.38      
          
s = standard deviation =    0.017161002      
          
Thus,          
          
z = (x - u) / s =    -1.165433114      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.165433114   ) =    0.12192186

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4.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.4      
u = mean =    0.38      
          
s = standard deviation =    0.017161002      
          
Thus,          
          
z = (x - u) / s =    1.165433114      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.165433114   ) =    0.12192186

Thus,

OPTION C: The result is usual because the probability that "p hat" is greater or equal to this sample proportion is greater than 5% [ANSWER, C]

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