Suppose that 43% of male Internet users aged 18 to 34 have visited an auction si

Question

Suppose that 43% of male Internet users aged 18 to 34 have visited an auction site at least once in the past month. If you interview 29 at random, what is the mean of the count X who have visited an auction site? What is the mean of the proportion p in your sample who have visited an auction site? Repeat the calculations in (a) for samples of size 290 and 2900. What happens to the mean count of successes as the sample size increases? mean count increases as n increases mean count remains constant as n increases mean count decreases as n increases What happens to the mean proportion of successes? mean proportion increases as n increases mean proportion remains constant as n increases mean proportion decreases as n increases

Explanation / Answer

(a)The mean of the count X who has visited the site is:n*p

=29*0.43=12.47

The mean of the proportion p^ is 0.43.

b)For n=290, the mean of the count X who has visited the site is: n*p

=290*0.43=124.7

The mean of the proportion p^ is 0.43.

For n=2900, the mean of the count X who has visited the site is: n*p

=2900*0.43=1247

The mean of the proportion p^ is 0.43.

c)The mean count increases as the sample size increases.

d)The mean proportion remains constant as n increases.

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