Question

A survey found that the average hotel room rate in Luzon is ₱2,300 and the average room rate in Mindanao is ₱2,150. Assume that the data were obtained from two samples of 75 hotels in Luzon and 80 hotels in Mindanao and the standard deviations were ₱250 and ₱215, respectively. At α=0.05, can it be concluded that there is a significant differences in the rates? (10 points)

Answer 1

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{250^2}{215^2}=1.3521$$

The critical values for $\alpha=0.05,$ $df_1=n_1-1=74$ degrees of freedom, $df_2=n_2-1=79$ degrees of freedom are $F_L = 0.6354$ and $F_U = 1.7622,$ and since $F = 1.3521$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

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

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

$$df=df_1+df_2=n_1+n_2-2=153$$

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df = 153.$

The critical value for this two-tailed test, $\alpha = 0.05, df=153$ degrees of freedom is $t_c =1.97559.$

The rejection region for this two-tailed test is $R = \{t: |t|> 1.97559\}.$

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

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{2300-2150}{\sqrt{\dfrac{(75-1)(250)^2+(80-1)(215)^2}{75+80-2}(\dfrac{1}{75}+\dfrac{1}{80})}}$$

$$=4.0125$$

Since it is observed that $t =4.0125>1.97559= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=153$ degrees of freedom, $t=4.0125$ is $p= 0.000094,$ and since $p= 0.000094<0.05=\alpha,$ it is concluded that the null hypothes is rejected.

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