he data set that I've chosen to work with in this week's discussion represents the percentage of unemployment in the state of Florida between the years of 2006 and 2017.

Year	Florida % Unemployment
2017	5.5
2016	6
2015	7
2014	8
2013	9.7
2012	11.5
2011	12.4
2010	13.3
2009	12.1
2008	7.5
2007	6.2
2006	5.5

The sample size (n) was found by counting the number of years that the percentage of unemployment in Florida was measured in the data. The sample mean () was found by computing "=average()" for the percentage of unemployment in google sheets. The standard deviation () was found by computing "=stdev.s()" for the percentage of unemployment, also in google sheets. For this data set, I've decided to use a 95% level of confidence. The alpha level () for a 95% confidence level is 0.05. Degrees of freedom (df) is equal to n-1. The margin of error () is calculated computing =confidence.t(alpha, standard deviation, sample size) in google sheets.

n= 12, = 8.725, = 2.925, = 0.05 or2= 0.025, df= 11, = 1.858

With these numbers we are able to determine the upper and lower confidence intervals of the sample mean.

= .

=(6.867,10.584)

The 95% confidence interval for the mean of the given data set is approximately [6.87,10.58]. From the data above, I've been able to conclude with 95% confidence that the true mean of the population from this sample lies within this confidence interval. The confidence interval and the standard deviation tell us that there is a relatively large amount of spread in the percentage of unemployment in Florida between 2006-2017. The variability is significant as the interval width and standard deviation show a noticeable spread from mean