**Margin**

### Question

**A poll taken this year asked 1012 adults whether they were fans of a particular sport and 75% said they were. Last year , 80% of a similar-size sample had reported being fans of the sport. Complete perts a through e below**
**a) Find the margin of error for the poll taken this year if one wants 90% confidence in the estimate of the percentage of adults who are fans of the sport.**

ME=Margin of Error

**b) Explain what That margin of error means.**

A. There is a 90% chance that the true proptrion of adults who are fans of the sport is within ME of this sample proportion

B. In 90% of samples of adults, the proportion who are fans of the sport will be within ME of the sample proportion.

C. One is 90% confident that this sam;le proportion is within ME of the true proportion of adults who are fans of the sport.

D. One is 90% ME confident that the true propotion of adults who are fans of the sport is equal to this sample proportion.
**C) If one wanted to be 80% confident instead of 90% confident, would the margin of error be larger of smaller?**

A. To be less xonfident the interval needs to contain the true proportion more often so the margin of error would be larger.

B. To be less confident the interval needs to contain the true proportion less often so the margin of error would be smaller.

C. To be less confident the interval needs to contain the true proportion more so the margin of error would be smaller.

D. To be less confident the interval needs to contain the true proportion less often so the margin of error would be larger.

**d) Find the margin of error for the poll taken this year if one wants 80% confidence in the estimate of the percent of adults who are fans of the sport.**

ME= margin of error

**e) In general, if all other aspects of the situation remain the same, will smaller margins of error produce greater or less confidence in the interval?**

1) Greater confidence

2) Less confidence

### Solutions..

- a) n=1012 , p=0.75

q=1-0.75

=0.25

mean= n^{*}q - 1012^{*}0.75 = 759

variance = n^{*}p^{*}q = 1012^{*}0.75^{*}0.25

=189.75

SD = sqrt(variance) = 13.77

Satandard Error(S.E) = SD/sqrt(1012)

= 0.43

critical value at 90% confindence interval is 1.64

Margin of Error(ME)= critical value^{*}Standard Error

= 1.64^{*}0.43

= 0.71

- b) One is 90% confident that the sample proportion is within +/- ME of the true portions of the adult who are fans of the sport.

- c)
** B **(To be less confident the interval needs to contain the true proportion less often so the margin of error would be smaller.)

- d) For 80% confidence interval the critical value is 1.28

ME = 1.28^{*}0.43

- = 0.55