Question

According to a survey performed by a real estate institute at USC, the sample mean rent in 2003 was $1,300 for Los Angeles County and $1,260 in Orange County. Assume the survey sampled 250 rents in each county. Also assume the standard deviation in rents was $350 in Los Angeles County and $450 in Orange County. a. Calculate a 90% confidence interval for the population mean rent for Los Angeles. Interpret.

b. Calculate a 90% confidence interval for the population mean rent for Orange County. Interpret.

c. Is there a good chance that the population mean rent in Orange County is higher than that in Los Angeles County? Explain.

Answer 1

(a) 90% confidence interval for the population mean rent for Los Angeles is [$1263.59 , $1336.41].

(b) 90% confidence interval for the population mean rent for Orange County is [$1213.18 , $1306.81].

(c) No, there is not a good chance that the population mean rent in Orange County is higher than that in Los Angeles County.

Step-by-step explanation:

We are given that according to a survey performed by a real estate institute at USC, the sample mean rent in 2003 was $1,300 for Los Angeles County and $1,260 in Orange County. Assume the survey sampled 250 rents in each county. Also assume the standard deviation in rents was $350 in Los Angeles County and $450 in Orange County.

(a) Firstly, the pivotal quantity for 90% confidence interval for the population mean rent for Los Angeles is given by;

     P.Q. =  $\frac{\bar X_1-\mu_1}{\frac{\sigma_1}{\sqrt{n_1} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean rent in 2003 for Los Angeles County = $1,300

             $\sigma_1$ = population standard deviation in Los Angeles County = $350

              $n_1$ = sample size for Los Angeles County = 250

              $\mu_1$ = population mean rent for Los Angeles

Here for constructing 90% confidence interval we have used One-sample z statistics because we know about population standard deviation.

So, 90% confidence interval for the population​ mean, $\mu_1$ is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90  {As the critical value of z at

                                  5% level of significance are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\bar X_1-\mu_1}{\frac{\sigma_1}{\sqrt{n_1} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$ < ${\bar X_1-\mu_1}$ < $1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$ ) = 0.90

P( $\bar X_1-1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$ < $\mu_1$ < $\bar X_1+1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$ ) = 0.90

90% confidence interval for $\mu_1$ = [ $\bar X_1-1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$ , $\bar X_1+1.6449 \times {\frac{\sigma_1}{\sqrt{n_1} } }$]

                                        = [ $1,300-1.6449 \times {\frac{\350}{\sqrt{250} } }$ , $1,300+1.6449 \times {\frac{\350}{\sqrt{250} } }$ ]

                                        = [1263.59 , 1336.41]

Therefore, 90% confidence interval for the population mean rent for Los Angeles is [$1263.59 , $1336.41].

This interval interprets that we are 90% confident that the population mean rent for Los Angeles will lie within this confidence interval i.e. between $1263.59 and $1336.41.

(b) Now, the pivotal quantity for 90% confidence interval for the population mean rent for Orange County is given by;

         P.Q. =  $\frac{\bar X_2-\mu_2}{\frac{\sigma_2}{\sqrt{n_2} } }$ ~ N(0,1)

where, $\bar X_2$ = sample mean rent in 2003 for Orange County = $1,260

             $\sigma_2$ = population standard deviation in Orange County = $450

             $n_2$ = sample size for Orange County = 250

             $\mu_2$ = population mean rent for Orange County

Here for constructing 90% confidence interval we have used One-sample z statistics because we know about population standard deviation.

So, 90% confidence interval for the population​ mean, $\mu_2$ is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90  {As the critical value of z at

                                  5% level of significance are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\bar X_2-\mu_2}{\frac{\sigma_2}{\sqrt{n_2} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$ < ${\bar X_2-\mu_2}$ < $1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$ ) = 0.90

P( $\bar X_2-1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$ < $\mu_2$ < $\bar X_2+1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$ ) = 0.90

90% confidence interval for $\mu_2$ = [ $\bar X_2-1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$ , $\bar X_2+1.6449 \times {\frac{\sigma_2}{\sqrt{n_2} } }$]

                                        = [ $1,260-1.6449 \times {\frac{\450}{\sqrt{250} } }$ , $1,260+1.6449 \times {\frac{\450}{\sqrt{250} } }$ ]

                                        = [1213.18 , 1306.81]

Therefore, 90% confidence interval for the population mean rent for Orange County is [$1213.18 , $1306.81].

This interval interprets that we are 90% confident that the population mean rent for Orange County will lie within this confidence interval i.e. between $1213.18 and $1306.81.

(c) No, there is not a good chance that the population mean rent in Orange County is higher than that in Los Angeles County because the Confidence interval for Los Angeles County has higher lower and upper limit of confidence interval as compared to that of Orange County.

That means there is higher chance of population mean rent to be higher in Los Angeles County.