Answer:
A 95% confidence interval for the true mean is [$3.39, $6.01].
Step-by-step explanation:
We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;
Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
                          P.Q.  =   ~
  ~ 
where,  = sample mean income =
 = sample mean income =  = $4.70
 = $4.70
             s = sample standard deviation =  = $1.83
  = $1.83
             n = sample of parking meters = 10
              = population mean
 = population mean 
<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>
<u>So, 95% confidence interval for the population mean, </u> <u> is ;</u>
<u> is ;</u>
P(-2.262 <  < 2.262) = 0.95  {As the critical value of t at 9 degrees of
 < 2.262) = 0.95  {As the critical value of t at 9 degrees of
                                             freedom are -2.262 & 2.262 with P = 2.5%}  
P(-2.262 <  < 2.262) = 0.95
 < 2.262) = 0.95
P(  <
 <  <
 <  ) = 0.95
 ) = 0.95
P(  <
 <  <
 <  ) = 0.95
 ) = 0.95
<u>95% confidence interval for</u>  = [
 = [  ,
 ,  ]
 ]
                                          = [  ,
 ,  ]
 ]
                                          = [$3.39, $6.01]
Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].
The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.
Also, the margin of error  =   
 
                                           =   = <u>1.31</u>
  = <u>1.31</u>