Zillow.com estimates of home values. Zillow.com is a real estate Web site that provides free estimates of the market value of homes. Refer to The Appraisal Journal (Winter 2010) study of the accuracy of Zillow’s estimates, Exercise 1.25 (p. 51). Data were collected for a sample of 2,045 single-family residential properties in Arlington, Texas. The researchers determined that Zillow overestimated by more than 10% the market value of 818 of the 2,045 homes. Suppose you want to estimate p, the true proportion of Arlington, Texas, homes with market values that are over-estimated by more than 10% by Zillow.

a. Find $\widehat{\mathbf{p}}$, the point estimate of p.

b. Describe the sampling distribution of .

c. Find a 95% confidence interval for p.

d. Give a practical interpretation of the confidence interval, part c.

e. Suppose a Zillow representative claims that p = .3. Is the claim believable? Explain.

Referring to The Appraisal Journal study of the accuracy of Zillow’s estimates, exercise 1.25. Sample of 2045 single-family residential properties in Arlington, Texas is taken.

Given the sample size is 2045 and the number of successes is 818.

The mean of the sampling distribution of the sample proportion$\widehat{p}$ is the population proportion p.

The sample proportion is an unbiased estimator of the population proportion.

Thus, the point estimate of the population proportion p is obtained below,

$\begin{array}{c}\widehat{\mathrm{p}}=\frac{\mathrm{Number}\mathrm{of}\mathrm{successes}\mathrm{in}\mathrm{the}\mathrm{sample}\left(\mathrm{x}\right)}{\mathrm{Sample}\mathrm{size}\left(\mathrm{n}\right)}\\ =\frac{818}{2045}\\ =0.4\end{array}$

Therefore, the point estimate$\widehat{p}$ of the population proportion p is 0.4.

The mean of the sampling distribution of$\widehat{p}$ is p.

$\widehat{p}$is an unbiased estimator of p.

The mean of the sampling distribution of$\widehat{p}$ is,

$\begin{array}{c}{\mu }_{\widehat{p}}=\widehat{p}\\ =0.4\end{array}$

The standard deviation of the sampling distribution of is,

$\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.4\times 0.6}{2045}}\\ =\sqrt{0.000117}\\ =0.0108\end{array}$

Therefore, the sampling distribution of $\widehat{p}$follows $N\left(0.4,0.0108\right)$.

Large-sample confidence interval for p is,

$\begin{array}{c}\left(\widehat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}}\right)=\left(\widehat{\mathrm{p}}-{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}},\widehat{\mathrm{p}}+{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}}\right)\\ =\left(0.4-\left(1.960\times 0.0108\right),0.4+\left(1.960\times 0.0108\right)\right)\left[\begin{array}{l}\mathrm{From}\mathrm{Standard}\\ \mathrm{Normal}\mathrm{Table}\end{array}\right]\\ =\left(0.4-0.0212,0.4+0.0212\right)\\ =\left(0.3788,0.4212\right)\end{array}$

Therefore, the 95% confidence interval for p is $\left(0.3788,0.4212\right)$.

There is a 95% confidence that the true proportion of market value overestimated by more than 10% by Zillow belongs to the interval $\left(0.3788,0.4212\right)$.

Here confidence interval is$\left(0.3788,0.4212\right)$ and the claim value of p is 0.3.

So, the value of p does not belong to the confidence interval $\left(0.3788,0.4212\right)$.

Therefore, at 5% level of significance it can be said that the claim is not believable.