Fair Market Rent. According to the document Out of Reach,...

Fair Market Rent. According to the document Out of Reach, published by the National Low Income Housing Coalition, the fair market rent (FMR) for a two-bedroom unit in Maine is $\$ 779 .$ A sample of 100 randomly selected two-bedroom units in Maine yielded the data on monthly rents, in dollars, given on the WeissStats CD. 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 twobedroom units in Maine is greater than the FMR of $\$ 779 ?$ 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.

Fair Market Rent. According to the document Out of Reach, published by the National Low Income Housing Coalition, the fair market rent (FMR) for a two-bedroom unit in Maine is $\$ 779 .$ A sample of 100 randomly selected two-bedroom units in Maine yielded the data on monthly rents, in dollars, given on the WeissStats CD. 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 twobedroom units in Maine is greater than the FMR of $\$ 779 ?$ 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.

Key Concepts

One-sample t-test

The one-sample t-test is a statistical method used to determine whether the mean of a sample significantly differs from a known or hypothesized population mean. It is particularly useful when the sample size is small and the population standard deviation is unknown, relying on the t-distribution to assess the probability of observing the sample mean under the null hypothesis.

Hypothesis Testing

Hypothesis testing involves formulating two opposing statements—the null hypothesis (typically stating no difference or effect) and the alternative hypothesis (indicating the presence of a difference or effect)—and then using sample data to assess whether there is enough evidence to reject the null hypothesis at a predefined significance level. In this context, the test examines whether the average rent exceeds the specified fair market rent.

Significance Level

The significance level, often denoted as alpha (?), is the threshold used in hypothesis testing to determine the probability of making a Type I error, which is the error of incorrectly rejecting a true null hypothesis. A 5% significance level means that there is a 5% risk of concluding that there is an effect or difference when there is none.

Outliers

Outliers are data points that differ significantly from other observations and can have a substantial influence on the results of statistical analyses, including the mean and standard deviation. An outlier can distort the outcome of a hypothesis test, so it is sometimes removed to assess the robustness of conclusions. Repeating the test after excluding outliers helps in understanding their impact on the overall analysis.

Conclusion Interpretation

Interpreting the conclusion of a hypothesis test involves balancing the statistical evidence with practical considerations. This includes assessing whether the evidence is strong enough to reject the null hypothesis after accounting for factors such as outliers and ensuring that the decision aligns with the broader context or research question. The conclusion conveys whether there is sufficient support to believe that the true mean differs from the hypothesized value.

Transcript

00:01 Talking about fair market rent with a sample size of 100 and a significance level of 5%.

00:06 The average is going to be calculated by saleming up all the values that we have divided by the sample size.

00:13 So that's going to be the sum divided by 100.

00:16 And we get 790 .96.

00:21 Now for the sample variance, s squared, that's simply going to be the standard deviation formula without the square root.

00:28 And we get that this is 8 ,072 .241.

00:36 We can take the square root of that and get our standard deviation.

00:44 And this turns out to be 89 .85.

00:52 Now for our null and alternative hypotheses, we find that the null hypothesis is that it is 779.

01:00 And for our alternative hypothesis, we find that the average is greater than 779.

01:08 Our t statistic is going to be 790 .96 minus 779 divided by the sample standard deviation of 89 .85, divided by the square root of 100.

01:26 This is approximately equal to 1 .331.

01:31 We have 99 degrees of freedom because of 100 minus 99, or 1.

01:37 Sorry.

01:40 And then the next thing we can do is find out whether we fail to reject or reject the null hypothesis.

01:49 So the closest degrees of freedom we can find in the student's graph is at degrees of freedom equals 95...

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