Course Project - Phase 4

Jessica Seifert

Rasmussen College

January 12, 2018

Question 1

Confidence intervals get used for giving a range of two figures whereby we can expect the population parameter which would include the mean to fall in within. Confidence intervals include confidence levels which get given by a percentage of how sure we are on where the population parameter will fall into the specified range. The confidence interval is calculated using the below formula: (x ̅-E<μ<x ̅+E).

A point estimate is represented as a single value and can also be said to be one statistic.

An example is the best point estimate for a population mean (μ) would be defined as a sample mean (x ̅). This is arguably the best point estimate because we are aware of the value of the entire population`s mean and therefore would take a sample of the population and calculate the sample mean and then use it on our confidence interval formula which would help in figuring the entire range of the whole population.

Confidence intervals get used as we are not aware of what is the real value of the population parameter. We, therefore, opt to use a small sample data to help us get a better idea of the data.

Question 2

After reviewing the data in the excel sheet, I found that the sample mean 62,306 although both sample mean and population mean are not the same it’s a reasonable point estimate for the population mean. I found the standard deviation of my spreadsheet is 19,149.21.

Question 3

We already know that the sample mean is (x ̅=62,306) and sample standard deviation (s=19,149.21), we would have to find our margin of error to construct our confidence interval. The formula to see our margin of error when σ is unknown is =t_(α/2) ∙s/√n. To solve this equating we need to find our t critical value corresponding to a 95% confidence level.

Step 1

Degrees of freedom (df) = n-1

364-1 = 363

Alpha (α) = 1-(confidence level/100)

= 1-(95/100)

=0.05

Critical probability (p) = 1-( α/2).

= 05/2 is .025

1-0.25 = p=.975.

Step 2

Use excel formula to find t critical value

=T.INV(.975,363) gives us t_(α/2)=1.967

To calculate margin of error

=1.967*19,149.21/sqrt(364)

E=1,974.261

(x ̅-E<μ<x ̅+E). 62,306 – 1,974.261=60,331.739,

and 62,306 + 1,974.261=64,280.261

The confidence interval in this scenario being (60,331.739 < μ < 64,280.261).

Question 4

This confidence interval means that the population mean of the salaries in Minnesota that range in between $40,000 and $120,000 have a 95% chance of being around $60,331.739 and $64,280.261 a year. The values give a range in which to expect the value of the mean of the population. The sample mean is the midpoint between the two numbers that provide the confidence interval. For this reason, the sample mean is referred to as the best point estimate of the population mean.

Question 5

The confidence intervals are 95% and 99%. This affects the standard deviation since the other factors are constant. As standard deviation changes so do risk, it reduces as one increases the confidence interval. Increase in confidence interval means a subsequent increase in standard deviation.

A confidence interval is meant to give a range of values where the estimated value will fall. The best value to use as the point of the estimate is the mean of the sample data. This occurs when the midpoint of two figures that are part of the range where the actual number could fall. By reducing the confidence interval, one increases the risk of the figure falling outside the set intervals. This means that the likelihood of obtaining the actual value would be reduced dramatically. The vice versa is also correct, increasing the confidence interval minimizes the probability of the value falling outside the set interval.