Question

In a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost five years earlier, in February 2008, only 35% of a similar-size sample had reported being baseball fans. a) Define p and p (what proportions are we looking at? This questions is not asking the value of p) b) Find the margin of error for the 2012 poll if we want 90% confidence in our estimate of the percent of national adults who are baseball fans. c) Explain what that margin of error means. d) If we wanted to be 99% confident would the margin of error be larger or smaller.e) Find that margin of error

Answer 1

Answer:

0.0259,0.0406

Step-by-step explanation:

Given that in a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost five years earlier, in February 2008, only 35% of a similar-size sample had reported being baseball fans.

a) p is the proportion of national adults who are baseball fans.

b) For 2012 poll, p=0.48

Std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.48*0.52}{1006} } \\=0.0158$

Margin of error at 90% = 1.645 * std error

= 0.0259

c) Margin of error is the admissible limit on either side of the mean where the sample mean can lie

d) Margin of error would be larger for 99% because critical value 2.58 >1.645

e) Margin of error 99% = $2.58*0.0157\\=0.0406$