The Township Board of Meridian Township wants to know how much public support there is for raising property taxes to fix and maintain the roads in the township. They randomly surveyed a group of residents and found that 61% are in favor of the tax increase. We want to calculate a 95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase. What is the upper bound for the 95% confidence interval if the estimated standard error of sample proportion is 0.0445? Give your answer to 4 decimal places.

95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is [0.5228 , 0.6972].

Step-by-step explanation:

We are given that the Township Board of Meridian Township randomly surveyed a group of residents and found that 61% are in favor of the tax increase.

The estimated standard error of sample proportion is 0.0445.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                          P.Q. =  ~ N(0,1)

where, = sample proportion of residents in favor of the tax increase = 61%

           n = sample of criminals

           p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.

The 95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is given by;

95% Confidence interval for p =  

Here, Standard error =   = 0.0445

And = significance level

So,   = 1.96

Hence, 95% Confidence interval for p =  

                                                           =  [ , ]

                                                           =  [0.5228 , 0.6972]

Therefore, 95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is [0.5228 , 0.6972].

Also, the upper bound for the 95% confidence interval is 0.6971.

sorry not sure

step-by-step explanation: