Question

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.

1. What percentage of the sample members believe the city should be spending more on public transportation?

2. What is the standard error of the proportion?

3. What is the margin of error for a 95% confidence interval for the population proportion?

4. What is the best interpretation for the 95% confidence interval?

5. Based on the findings, is it safe to say that a majority of the city's population wants more spent on public transportation?

Answer 1

1)

Number of Items of Interest,   x =   262
Sample Size,   n =    400
      
Sample Proportion ,    p̂ = x/n =    0.6550

2) Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.023768

3)α=0.05

z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  

margin of error , E = Z*SE =    1.960   *   0.02377   =   0.0466

4)

95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.65500   -   0.04659   =   0.60841
Interval Upper Limit = p̂ + E =   0.65500   +   0.04659   =   0.70159
                  
95%   confidence interval is (   0.608   < p <    0.702   )

5)

yes , it is safe to say that a majority of the city's population wants more spent on public transportation because both ends of CI is more than 0.50