Question

According to a survey in a​ country, 39​% of adults do not own a credit card....

According to a survey in a​ country, 39​% of adults do not own a credit card. Suppose a simple random sample of 300 adults is obtained. Complete parts​ (a) through​ (d) below. ​(a) Describe the sampling distribution of ModifyingAbove p with caret​, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of ModifyingAbove p with caret below. A. Not normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10 B. Not normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10 C. Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10 D. Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10 Determine the mean of the sampling distribution of ModifyingAbove p with caret. mu Subscript ModifyingAbove p with caret Baseline equals nothing ​(Round to two decimal places as​ needed.) Determine the standard deviation of the sampling distribution of ModifyingAbove p with caret. sigma Subscript ModifyingAbove p with caretequals nothing ​(Round to three decimal places as​ needed.) ​(b) What is the probability that in a random sample of 300 ​adults, more than 41​% do not own a credit​ card? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. If 100 different random samples of 300 adults were​ obtained, one would expect nothing to result in more than 41​% not owning a credit card. ​(Round to the nearest integer as​ needed.) ​(c) What is the probability that in a random sample of 300 ​adults, between 36​% and 41​% do not own a credit​ card? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. If 100 different random samples of 300 adults were​ obtained, one would expect nothing to result in between 36​% and 41​% not owning a credit card. ​(Round to the nearest integer as​ needed.) ​(d) Would it be unusual for a random sample of 300 adults to result in 108 or fewer who do not own a credit​ card? Why? Select the correct choice below and fill in the answer box to complete your choice. ​(Round to four decimal places as​ needed.) A. The result is not unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is nothing​, which is less than​ 5%. B. The result is unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is nothing​, which is less than​ 5%. C. The result is unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is nothing​, which is greater than​ 5%. D. The result is not unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is nothing​, which is greater than​ 5%. Click to select your answer(s).

Homework Answers

Answer #1

p = 0.39

n = 300

a) Approximately normal, because n < 0.05N and np(1 - p) > 10.

= 0.39

= sqrt(p(1 - p)/n)

     = sqrt(0.39 * (1 - 0.39)/300)

     = 0.028

b) P( > 0.41)

= P(( - )/> (0.41 - )/)

= P(Z > (0.41 - 0.39)/0.028)

= P(Z > 0.71)

= 1 - P(Z < 0.71)

= 1 - 0.7611

= 0.2389

c) P(0.36 < < 0.41)

= ((0.36 - )/ < ( - )/ < (0.41 - )/)

= P((0.36 - 0.39)/0.028 < Z < (0.41 - 0.39)/0.028)

= P(-1.07 < Z < 0.71)

= P(Z < 0.71) - P(Z < -1.07)

= 0.7611 - 0.1423

= 0.6188

d) P(< 0.36)

= P(( - )/< (0.36 - )/)

= P(Z < (0.36 - 0.39)/0.028)

= P(Z < -1.07)

= 0.1423

The result is not unusual because the probability that with caret is less than or equal to the sample proportion is 0.1423 which is greater than 5%.

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