A poll taken this year asked 1035 adults whether they were fans of a particular
ID: 2908437 • Letter: A
Question
A poll taken this year asked
1035
adults whether they were fans of a particular sport and
29?%
said they were. Last? year,
30?%
of a? similar-size sample had reported being fans of the sport. Complete parts a through e below.
Summer 201 e Profes(1) 7/29/18 6:05 PM Quiz: Quiz:Module 4 Submit Quiz This Question: 1 pt 2 of 10 (0 complete) ? This Quiz: 10 pts possible A poll taken this year asked 1035 adults whether they were fans o a particular sport and 29 6 said they ere Last year, 30% of a similar size sample had reported being fans of the sport. Complete parts a through e bel a) Find the margin of error for the poll taken this year if one wants 95% confidence in the estimate of the percentage of adults who are fans of the sport. ME-(Round to three decimal places as needed.) b) Explain what that margin of error means. ( A. One is 95% confident that this sample proportion is within ± ME of the true proportion of adults who are fans of the sport. ME ofthe sample proportion. B. C. D. In 95% of samples of adults, the proportion who are fans of the sport will be within There is a 95% chance that the true proportion of adults who are fans ofthe sport is within?ME of this sample proportion. One is 95% ± ME confident that the true proportion of adults who are fans of the sport is equal to this sample proportion. c) lf one wanted to be 85% confident instead of 95% confident, would the margin of error be larger or smaller? O A. O B. ° C. To be less confident, the interval needs to contain the true proportion less often, so the margin of error would be smaller To be less confident, the interval needs to contain the true proportion more often, so the margin of error would be smaller To be less confident, the interval needs to contain the true proportion more often, so the margin of error would be larger 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 85% confidence in the estimate of the percent of adults who are fans of the sport ME(Round to three decimal places as needed.) 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? O Less confidence O Greater confidence Click to select your answer(s).Explanation / Answer
Answer (a)
Given
Sample Size: n = 1035
Sample Proportion: p? = 0.29
Level of Confidence (%) = 95
Formula for Margin of Error is z?/2•?p?(1 - p?)/n
Level of Confidence = 95%
? = 100% - (Level of Confidence) = 5%
?/2 = 2.5% = 0.025
Calculate z?/2 by using standard normal distribution with ?/2 = 0.025 as right-tailed area and left-tailed area.
z?/2 = 1.96 (Obtained using z distribution table for ?/2 = 0.025)
Margin of Error = z?/2•?p?(1 - p?)/n = 1.96•?0.29(1 - 0.29)/1035
Margin of Error = 0.028
Answer (b)
Option A is correct
Reason: A margin of error tells you how many percentage points your results will differ from the real population value.
Answer c)
Option A is correct
Reason: If you want more confidence that an interval contains the true parameter, then the intervals will be wider. Similarly, if you want less confidence that n interval contains the true parameter, then the intervals will be narrower. For example, If you want to be 100.000% sure that an interval contains the true population, it has to contain every possible value so be very wide. If you are willing to be only 50% sure that an interval contains the true value, then it can be much narrower.
Answer d)
Given
Sample Size: n = 1035
Sample Proportion: p? = 0.29
Level of Confidence (%) = 95
Formula for Margin of Error is z?/2•?p?(1 - p?)/n
Level of Confidence = 85%
? = 100% - (Level of Confidence) = 15%
?/2 = 7.5% = 0.075
Calculate z?/2 by using standard normal distribution with ?/2 = 0.075 as right-tailed area and left-tailed area.
z?/2 = 1.44 (Obtained using z distribution table)
Margin of Error = z?/2•?p?(1 - p?)/n = 1.44•?0.29(1 - 0.29)/1035
Margin of Error = 0.020
Answer (e)
Less Confidence
Reason: Same as discussed in Answer (c).