Answer to Question #250287 in Statistics and Probability for smilynne

Question

A real estate agent compares the selling price of townhouses in two major cities in National Capital Region to see if there is a difference in price. Is there enough evidence to reject the claim that the average price of a townhouse in Quezon City is higher than Makati City? Use 0.05 alpha level.

 Quezon City Makati City

_ _

X1 = 2,140, 000 X2 = 1,970,000

𝒔𝟏  = 226,000 𝒔2 = 243,000

𝒏1 =  47 𝒏2 = 45

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq \mu_2$

​$H_1:\mu_1>\mu_2$

​This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.05.$

The degrees of freedom are computed as follows, assuming that the population variances are unequal:

$df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}$

$=\dfrac{(\dfrac{226^2}{47}+\dfrac{243^2}{45})^2}{\dfrac{(226^2/47)^2}{47-1}+\dfrac{(243^2/45)^2}{45-1}}\approx88.80012155$

It is found that the critical value for this right-tailed test is $t_c = 1.6622,$ for $\alpha = 0.05$

and $df = 88.80012155.$

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

$t = \dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}=\dfrac{2140-1970}{\sqrt{\dfrac{226^2}{47}+\dfrac{243^2}{45}}}\approx3.470889$

Since it is observed that $t = 3.470889 > 1.6622=t_c ,$ it is then concluded that the null hypothesis is rejected.

​Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=88.80012155,$ $t=3.470889$ is $p =0.000402,$ and since $p = 0.000402 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

​

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$df=n_1-1+n_2-1=47-1+45-1=90$

It is found that the critical value for this right-tailed test is $t_c = 1.661961,$ for $\alpha = 0.05$

and $df = 90.$

The rejection region for this right-tailed test is $R = \{t: t > 1.661961\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$ $= \dfrac{2140-1970}{\sqrt{\dfrac{(47-1)(223)^2+(45-1)(246)^2}{47+45-2}(\dfrac{1}{47}+\dfrac{1}{45})}}$

$\approx3.475509$

Since it is observed that $t = 3.475509 > 1.661961=t_c ,$ it is then concluded that the null hypothesis is rejected.

​Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=90,$ $t=3.475509$ is $p =0.000393,$ and since $p = 0.000393 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than$\mu_2,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the average price of a townhouse in Quezon City is higher than Makati City, at the $\alpha = 0.05$ significance level.