Statistical Data

2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

(21) You took a sample of size 21 from a normal distribution with a known standard deviation, . In order to find a 90% confidence interval for the mean, You need to find.

The mean for the column “mean” is 3.56. It is very close to the parameter of interest but is not equal to it. You can calculate a confidence interval for the mean of the mean column, but a specific confidence interval would need to be provided. In that case, the confidence interval would be centered on 3.56, not 3.5.

A business wants to estimate the true mean annual income of its customers.  It randomly samples 220 of its customers.  The mean annual income was $61,400 with a standard deviation of $2,200.  Find a 95% confidence interval for the true mean annual income of the business’ customers.

b. Construct a 91.5% confidence interval for the mean time it take for all workers who are employed in downtown Toronto

Standard Deviation for the mean column is 0.476Standard Deviation for the median column is 0.754Standard deviation for the mean column has least variability

B. 95% confidence intervals for the proportion of all students, the proportion of all male students, and the proportion of all female students who were involved in copying off the Internet is provided in the following table.

Below is a table and scatter plot displaying David Ortiz’s home runs earned during the past five years with the Boston Red Sox. The data collected is based off of David Ortiz’s home runs earned over the course of that correlating baseball season. The table organizes data into the amount of times David Ortiz was at bat, the amount of earned home runs, as well as the percentage of hits that resulted in home runs. In addition to the table, summary statistics were created to show the mean, variance, standard deviation as well as median of earned home runs. These values show that David Ortiz has been consistent with home runs earned with little variance.

The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.

I have determined that the 95% Lower Confidence Bound is 42.61 and the for actual population mean is greater than 42.61 with a probability 0.95.

The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.

In this project we were given the case of customer complaints that the bottles of the brand of soda produced in our company contained less than the advertised sixteen ounces of product. Our boss wants us to solve the problem at hand and has asked me to investigate. I have asked my employees to pull Thirty (30) bottles off the line at random from all the shifts at the bottling plant. The first step in solving this problem is to calculate the mean (x bar), the median (mu), and the standard deviation (s) of the sample. All of those calculations were easily computed in excel. The mean was computed by entering:

Since 3.27 the t statistic is in the rejection area to the right of =1.701, the level of