Question

A study was made comparing the cost of a one-bedroom apartment in philadelphia with the cost of similar apartments in Baltimore. A sample of 30 apartments in Philadelphia showed a sample mean of $950 with a standard devation of $50. A sample of 25 apartments in Baltimore showed a sample mean of $915 and a sample stand deviation of $45. Test to see if there is a significant difference in mean rental rate between the two cities. Use a 5% leve of significance.What is your conclusion?

Answer 1

Answer:

$t=\frac{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73$  

$p_v =2*P(t_{53}>2.73) =0.0086$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Philadelphia) is  significantly different than the mean for the group 2 (Baltimore).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: $\mu_1 = \mu_2$  

Alternative hypothesis: $\mu_1 \neq \mu_2$  

Or equivalently:  

Null hypothesis: $\mu_1 - \mu_2 = 0$  

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$  

Our notation on this case :  

$n_1 =30$ represent the sample size for group 1  (Philadelphia)

$n_2 =25$ represent the sample size for group 2  (Baltimore)

$\bar X_1 =950$ represent the sample mean for the group 1  

$\bar X_2 =915$ represent the sample mean for the group 2  

$s_1=50$ represent the sample standard deviation for group 1  

$s_2=45$ represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Critical values

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution $\alpha/2 = 0.025$ of the area.  

The distribution on this case since we don't know the population deviation for both samples is the t distribution with $df=n_1+n_2 -2= 30+25-2=53$ degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,53)", "=T.INV(1-0.025,53)"

And we got: (-2.01, 2.01)

Calculate th statistic

The statistic is given by this formula:  

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$  

And now we can calculate the statistic:  

$t=\frac{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73$  

The degrees of freedom are given by:  

$df=30+25-2=53$

P value

And now we can calculate the p value using the altenative hypothesis:  

$p_v =2*P(t_{53}>2.73) =0.0086$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Philadelphia) is  significantly different than the mean for the group 2 (Baltimore).