2.5pts] A mayor of a big city finds that 262 of 400 randomly selected city resid
ID: 3042358 • Letter: 2
Question
2.5pts] A mayor of a big city finds that 262 of 400 randomly selected city residents believe that the city should spend more money on public transportation. Use this information for a large-sample inference of the proportion of all city residents who believe that the city should be spending more.
What percentage of the sample members believe the city should be spending more?
Calculate the standard error of the proportion.
Find and interpret the 95% confidence interval for the population proportion.
Based on your findings, is it safe to say that a majority of the city’s population wants more spent on public transportation? Explain your answer.
Explanation / Answer
Answer to the question is as follows. Please write back in case you have doubts:
1. The percentage of the sample members believe the city should be spending more
= Residents who believe that the city should spend more money on public transportation/ total people in sample
= 262/400
= .655
Standard error in proportion, SE
= sqrt( p (1-p)/ n)
= sqrt( .655*.345/400)
= .0238
The 95% CI is given by :
p^ +/- Z*SE
= 0.655 +/- 1.96*.0238
= 0.6084 to 0.7017
It means that if repeated samples were taken and the 95%confidence interval was computed for each sample, 95% of the intervals would contain the population mean within this range of 0.6084 to 0.7017
Since this proportoin is more than .05 ( both upper and lower bounds are more than .5) we can safely say that majority of the city' population wants more spent on public expenditure.
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