# Answer in Statistics and Probability for smilynne #250287

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

= 226,000 2 = 243,000

1 =  47 2 = 45

The following null and alternative hypotheses need to be tested:

“H_0:mu_1leq 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.

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