STATISTICS - Lab #6
Statistical Concepts: Data Simulation Discrete Probability Distribution Confidence Intervals
Calculations for a set of variables
Open the class survey results that were entered into the MINITAB worksheet.
We want to calculate the mean for the 10 rolls of the die for each student in the class. Label the column next to die10 in the Worksheet with the word mean. Pull up Calc > Row Statistics and select the radio-button corresponding to Mean. For Input variables: enter all 10 rows of the die data. Go to the Store result in: and select the mean column. Click OK and the mean for each observation will show up in the Worksheet.
We also want to calculate the median for the 10 rolls of the die. Label the
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Either show work or explain how your answer was calculated.
Mean: Summation xP(x) = 1(1⁄6) +2(1⁄6) + 3(1⁄6) + 4(1⁄6) + 5(1⁄6) +6(1⁄6) = 21⁄6= μ 3.5
Standard deviation: sq. root ((1-3.5)^2 •(1⁄6) + (2-3.5)^2•(1⁄6) + (3-3.5)^2•(1⁄6) + (4-3.5)^2•(1⁄6) + (5-3.5)^2•(1⁄6) + (6-3.5)^2•(1⁄6))= sq. root2.916=σ 1.707
3.) Give the mean for the mean column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Mean: 3.560. Yes, this is very closely centered around the parameter of interest (3.5)
4.) Give the mean for the median column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Median: 3.600. Yes, this too is also centered around the parameter of interest (3.5).
5.) Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
σ of Mean: .0476 σ of Median: .0754
The standard deviation of the Means is smaller, thus having less variability than the Median, meaning the data for the Means is grouped closer
2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?
So, we have a distribution with a mean of 20,000 and a standard deviation of 5,102.
b) In order to calculate the mean or average for the governors and CEO’s, I added together all the figures and divided that sum by 4 since there
(b) The value of Z with an area of 5% in the right tail, but not the sample mean.
4. Calculate the following measures of central tendency for the set of cube measurement data. Show your work or explain your procedure for each.
4. Give the mean for the median column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)
2. Compute the means for the following set of scores saved as Ch. 2 Data Set 3 using IBM® SPSS® software. Print out a copy of the output. (Please refer to attachment)
Let’s assume you have taken 1000 samples of size 64 each from a normally distributed population. Calculate the standard deviation of the sample means if the population’s variance is 49.
5. Find the sample variance s2 for the following sample data. Round your answer to the nearest hundredth.
2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.
A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are the results of his random sampling. Can he conclude, at the .10 level of significance, that
Find the mean, median, SD & IQR for the data in (1) after it has been transformed
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:
Based on the chart, the mean was calculated by adding up the sum of the list and divide 18, which the number of the total listed prices. The mean is 135,000, which mean the average of the listed price. Secondly, the median was calculated by listing the number in numerical order from lowest to highest and located the number in the middle 126,000. The median represents the middle number of the listed price. After calculating the median I located the minimum and maximum based the lowest and highest data, which are 48,000 and 338,000. These represent the range of the listed price. Lastly, I used the formula to get the
The median is obtained as the middle value of the two central values (the 25th and the 26th):