## Choosing a Statistical Test

**Example: SD = 1.23**

## Computing Central Tendency

## Hypothesis Testing

## Solution to Assignment

**Scenario:
**

A community leader in a small rural county in Alabama read an article “67 Alabama Counties, Ranked from Most to Least-educated” published by AL.Com (2017). After reading this article, he became interested in determining two things: 1) an overview of education in the Alabama counties and 2) how well Alabama counties compared to the national average of college level educated adults. National average = 33%.

The mean percent of adults with a college degree or higher in Alabama is 16.65%

The median percent of adults with a college degree or higher in Alabama 14.10%

The lowest percent of adults with a college degree or higher in Alabama is 8.20% while the country with the lowest percent is Conecub

The highest percent of adults with a college degree or higher in Alabama is 40.80% while the country with the lowest percent is Shelby

The range value is 32.60

The standard deviation is 7.04

The skewness statistic is 1.68. If the skewness is less than -1 or greater than 1, the data are highly skewed. Therefore, since the skewness statistic is greater than 1, we say that the data is highly skewed.

The values of the 30th, 60th, and 90th percentiles are 12.44, 15.54, and 28.60 respectively. Percentiles tell you how a value compares to other values. The 30th percentile means 12.44% in adults with a college degree or higher in Alabama contribute about 30% of the education percentage.

The appropriate statistical test to show how well Alabama counties compared to the national average of college level educated adults is the One-sample t-test

Null Hypothesis (H0): there is no statistical difference in the sample mean and the hypothesized mean (33.00) i.e. H_0: μ≠μ_0

Alternative Hypothesis (H0): there is statistical difference in the sample mean and the hypothesized mean (33.00) i.e. H_0: μ=μ_0

Alternative Hypothesis (H0): there is statistical difference in the sample mean and the hypothesized mean (33.00) i.e. H_0: μ>μ_0

The computed value of the test statistics is -18.996

The value of the degrees of freedom reported in the output is 66

The reported level of significance is 0.000

Based on the output, the null hypothesis will be rejected as the test is statistically significant because the level of significance is less than 0.05. Therefore, we conclude that there is statistical difference between the sample mean and the hypothesized mean (33.00).

The reported confidence interval is (-18.06, -14.63). This shows that the researcher is 95% confident that the population parameter is between -18.06 and -14.63.