Using p-hat formula and probability,
we get
a) Sampling distribution of p-hat = 0.02029286
b)The probability that in a random sample of 500 adults more than 30% do not own a credit card is
=0.3121
c) The probability that in a random sample of 500 adults between 25% and 30% do not own a credit card is 0.6635
We have given that
29% of adults do not own a credit card according to creditcard.com.
a) Sample size (n) = 500 of adults have own credit card
sample proportion, for the adults do not own a credit= 29% = 0.29
Normal distribution exits here with p
= 1-0.29=0.71 and
Standard error (p-hat) =sqrt(p(1-p)/n)
=sqrt(0.71× 0.29/500)
p-hat = 0.02029286
b) Now, more than 30% do not have own credit card.
P(p-hat>0.3) = P((phat- p)/sqrt(p×(1-p)/n) >(0.3-0.29)/sqrt(0.29× (1-0.29)/500))
=P(Z>0.49) =0.3121 (from standard normal table)
c) Sample size (n) = 500 and p-hat lies between 25% to 30% .
P(0.25<phat<0.3)
= P((0.25-0.29)/sqrt(0.29*(1-0.29)/500) <Z< (0.3-0.29)/sqrt(0.29*(1-0.29)/500))
=P(-1.97<Z<0.49)
=0.6635 (from standard normal table)
Hence, we calculated all the required probabilities.
a) p-hat = 0.02029286
b) P(p-hat>0.3) =0.3121
c) P(0.25<phat<0.3) =0.6635
To learn more about Standard error or p-hat , refer:
https://brainly.com/question/12360461
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Complete question:
According to creditcard.com, 29% of adults do not own a credit card.
a. Suppose a random sample of 500 adults is asked, "Do you own a credit card?" Describe the sampling distribution of p-hat, the proportion of adults who own a credit card.
b. What is the probability that in a random sample of 500 adults more than 30% do not own a credit card?
c. What is the probability that in a random sample of 500 adults between 25% and 30% do not own a credit card?
