Question

Fair Market Rent. According to the document Out of Reach published by the National Low Income Housing Coalition, the fai market rent (FMR) for a two-bedroom unit in the United States is 949 A sample of 100 randomly selected two-bedroom units yielded the data on monthly rents, in dollars, given on the WeissStats site. Use the technology of your choice to do the following.

a. At the 5 % significance level, do the data provide sufficient evidence to conclude that the mean monthly rent for two-bedroom units is greater than the FMR of $949? Apply the one-mean t-test.

b. Remove the outlier from the data and repeat the hypothesis test in part (a).

c. Comment on the effect that removing the outlier has on the hypothesis test.

d. State your conclusion regarding the hypothesis test and explain your answer.

Short answer

a) The data does not support the conclusion that the average monthly rent for two-bedroom homes is higher than the FMR of $949.

b) The results support the conclusion that the average monthly rent for two-bedroom flats is higher than the FMR of $949.

c) $4.8 (=89.85-85.05)

Step-by-step solution

Subpart (a) Step 1:

Check whether the data provide sufficient evidence to conclude that the mean monthly rent for two-bedroom units is greater than the FMR of $ 949.

State the null and alternative hypothesis:

Null hypothesis:

$H_0:\mu =\$949$

That is, the mean monthly rent for two-bedroom units is not greater than the FMR of $\$949$.

Alternative hypothesis:

$H_a:\mu >949$

That is, the mean monthly rent for two-bedroom units is greater than the FMR of $\$949$.

Subpart (a) Step 2:

Compute the value of the test statistic by using MINITAB.

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample t.

Step 2: In Samples in Column, enter the column of Rent.

Step 3: In Perform hypothesis test, enter the test mean as 949.

Step 4: Increase the Confidence Level to 95.0 if desired.

Step 5: As an option, select greater than.

Step 6: In all dialogue boxes, click OK.

Subpart (a) Step 3:

MINITAB Output:

One-Sample T: RENT

Test of $\mu =949vs>949$

Variable	N	Mean	StDev	SE Mean	95% Lower Bound	T	P
RENT	100	960.96	89.85	8.98	946.04	1.33	0.093

From the output, the value of test statistic is $1.33$, and the P-value is$0.093$

Subpart (a) Step 4:

P - value method:

Rule of rejection:

If the null hypothesis is true, then reject it.

Conclusion:

Here, the P-value is 0.093 which is greater than the level of significance. That is, $P(=0.093)>\alpha (=0.05)$.

Therefore, the null hypothesis is not rejected at $5\%$level.

Thus, it can be conclude that the test results are not statistically significant at $5\%$level of significance.

Subpart (a) Step 5: 

Interpretation:

The data do not provide sufficient evidence to conclude that the mean monthly rent for two bedroom units is greater than the FMR of $ 949 .

Subpart (b) Step 1:

Construct the boxplot by using

MINITAB procedure:

Step 1: Choose Graph > Boxplot or Stat > EDA > Boxplot.

Step 2: Select Simple under Multiple Y's. Select OK.

Step 3: In Graph variables, enter the data of Rent.

Step 4: Click OK.

MINITAB Output:

From the box plot, the observation 662 is considered as outlier. Moreover, the distribution of the data is slightly skewed.

Subpart (b) Step 2:

Repeat the hypothesis test in part (a) after removing the outlier from the data.

Compute the value of the test statistic by using MINITAB.

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample t

Step 2: In Samples in Column, enter the column of Rent.

Step 3: In Perform hypothesis test, enter the test mean as 949 .

Step 4: Go to Options and choose Confidence Level 95.0.

Step 5: Select the greater than option.

Step 6: In all dialogue boxes, click OK.

MINITAB Output:

One-Sample T: RENT

Test of $\mu =949vs>949$

Variable	N	Mean	StDev	SE Mean	95% Lower Bound	T	P
RENT	99	963.98	85.05	8.55	949.79	1.75	0.041

from the output, the value of test statistic is 1.75 and the p-value is 0.041

Subpart (b) Step 3:

P - value approach:

Here, the P-value is 0.041 which is lesser than the level of significance. That is,

$P(=0.041)<\alpha (=0.05)\text{.}$

Therefore, the null hypothesis is rejected at 5 % level.

As a consequence, the test findings can be concluded to be statistically significant at the 5% level of significance. As a result, the results support the conclusion that the average monthly rent for two bedroom homes is higher than the FMR of $949.

Subpart (c) Step 1:

Comments on the effect that removing the outlier has on the hypothesis test.

From part b, there is a rejection of the null hypothesis when removing the outlier because the sample mean is increased by $\$3.02(=963.98-960.96)$and the sample standard deviation decreased by $\$4.8(=89.85-85.05)$.

Subpart (d) Step 1:

Explanation:

Even though the outlier's reaction is substantial, the sample size is considerable in both circumstances. If the outliers are not eliminated from the data, the hypothesis test results will not generate accurate population findings.

When outliers are eliminated from the data, the hypothesis test produces reliable population findings.