91影视

Referring to The Appraisal Journal study of the accuracy of Zillow鈥檚 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,

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

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,

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

The standard deviation of the sampling distribution of is,

$\begin{array}{c}{蟽}_{\widehat{p}}=\sqrt{\frac{p\left(1鈭�p\right)}{n}}\\ =\sqrt{\frac{0.4脳0.6}{2045}}\\ =\sqrt{0.000117}\\ =0.0108\end{array}$

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

Large-sample confidence interval for p is,

$\begin{array}{c}\left(\widehat{\mathrm{p}}卤{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}}\right)=\left(\widehat{\mathrm{p}}鈭�{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}},\widehat{\mathrm{p}}+{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\widehat{\mathrm{p}}\widehat{\mathrm{q}}}{\mathrm{n}}}\right)\\ =\left(0.4鈭�\left(1.960脳0.0108\right),0.4+\left(1.960脳0.0108\right)\right)\text{鈥夆赌�}\left[\begin{array}{l}\mathrm{From}\text{鈥夆赌�}\mathrm{Standard}\\ \mathrm{Normal}\text{鈥夆赌�}\mathrm{Table}\end{array}\right]\\ =\left(0.4鈭�0.0212,0.4+0.0212\right)\\ =\left(0.3788,0.4212\right)\end{array}$

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.