According to creditcardcom 29 of adults do not own a credit

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 of p?, the proportion of adutls 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% -30% do not own a credit card?

d) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Why?

Solution

a) Suppose a random sample of 500 adults is asked, \"Do you own a credit card?\" Describe the sampling of p?, the proportion of adutls who own a credit card.

It follows Normal distribution with p=1-0.29=0.71 and

standard deviatoin =sqrt(p*(1-p)/n)

=sqrt(0.71*0.29/500)

=0.02029286

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b) What is the probability that in a random sample of 500 adults more than 30% do not own a credit card?

P(phat>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)

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c) What is the probability that in a random sample of 500 adults between 25% -30% do not own a credit card?

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)

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d) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Why?

125/500 =0.25

P(phat<0.25) =P(Z<-1.97) = 0.0244 (from standard normal table)

Since the probability is less than 0.05, it would be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card