Question

A realtor wants to estimate the rean af houses in Mssion Calfamia. She colects a of 38 recent housesales in $1000s), a of which sown in the able. Assume that the standard deviaton is 100 House P 520 480 521 533 a. Construct 95% and 98% confidence intervals for the mean price of all houses in Mission Viejo, CA Use rounded sample mean. Do not round intermediate calculations. Round "z" value to 3 decimal places and final answers to 2 decimal places) 370 430 Level 95⅔ 98% Confidence Interval to to 399 560 425 669 540 588 445 735 537

Answer 1

From the data, the mean $\bar{x}$ = 516.0278

Given, $\sigma$ = 100

The 95% Confidence Interval

Since population standard deviation is known and n > 30, the Zcritical (2 tail) for $\alpha$ = 0.05, is 1.960

The Confidence Interval is given by $\bar{x}$ $\pm$ ME, where

$ME = Zcritical * \frac{s}{\sqrt{n}} = 1.960 * \frac{100}{\sqrt{36}} = 32.6667$

The Lower Limit = 516.0278 - 32.6667 = 483.3611 $\approx$     483.36

The Upper Limit = 516.0278 + 32.6667 = 548.6945   $\approx$    548.69

The 95% Confidence Interval is 483.36 to 548.69

_________________________________________________________

The 99% Confidence Interval

Since population standard deviation is known and n > 30, the Zcritical (2 tail) for $\alpha$ = 0.01, is 2.576

The Confidence Interval is given by $\bar{x}$ $\pm$ ME, where

$ME = Zcritical * \frac{s}{\sqrt{n}} = 1.960 * \frac{100}{\sqrt{36}} = 42.9333$

The Lower Limit = 516.0278 - 42.9333 = 473.0945 $\approx$     473.09

The Upper Limit = 516.0278 + 42.9333 = 558.9611 $\approx$    558.96

The 95% Confidence Interval is 473.09 to 558.96

_________________________________________________________

Confidence level	Confidence Interval
95%	483.36	to	548.69
99%	473.09	to	558.96