MATH221_W4_Lab Dondra Houston
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MATH221_W4_Lab Dondra Houston

Course Number: ECON 101, Spring 2011

College/University: UNLV

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Statistics Lab Week 4 Name:__Dondra Houston_____________________ MATH221 Statistical Concepts: Probability Binomial Probability Distribution Calculating Binomial Probabilities > Open a new MINITAB worksheet. > We are interested in a binomial experiment with 10 trials. First, we will make the probability of a success . Use MINITAB to calculate the probabilities for this distribution. In column C1 enter...

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Lab Statistics Week 4 Name:__Dondra Houston_____________________ MATH221 Statistical Concepts: Probability Binomial Probability Distribution Calculating Binomial Probabilities > Open a new MINITAB worksheet. > We are interested in a binomial experiment with 10 trials. First, we will make the probability of a success . Use MINITAB to calculate the probabilities for this distribution. In column C1 enter the word success as the variable name (in the shaded cell above row 1. Now in that same column, enter the numbers zero through ten to represent all possibilities for the number of successes. These numbers will end up in rows 1 through 11 in that first column. In column C2 enter the words one fourth as the variable name. Pull up Calc > Probability Distributions > Binomial and select the radio button that corresponds to Probability. Enter 10 for the Number of trials: and enter 0.25 for the Event probability:. For the Input column: select success and for the Optional storage: select one fourth. Click the button OK and the probabilities will be displayed in the Worksheet. > Now we will change the probability of a success to . In column C3 enter the words one half as the variable name. Use similar steps to that given above in order to calculate the probabilities for this column. The only difference is in Event probability: use 0.5. > Finally, we will change the probability of a success to . In column C4 enter the words three fourths as the variable name. Again, use similar steps to that given above in order to calculate the probabilities for this column. The only difference is in Event probability: use 0.75. Plotting the Binomial Probabilities 1. Create plots for the three binomial distributions above. Select Graph > Scatter Plot and Simple then for graph 1 set Y equal to one fourth and X to success by clicking on the variable name and using the select button below the list of variables. Do this two more times and for graph 2 set Y equal to one half and X to success, and for graph 3 set Y equal to three fourths and X to success. Paste those three scatter plots below. Sca t t e r pl ot of one f our t hv sSucce ss 0.30 0.25 one f ou r t h 0.20 0.15 0.10 0.05 0.00 0 2 4 Succe ss 6 8 10 8 10 Sca t t e r pl ot of one ha l f v sSucce ss 0.25 one ha lf 0.20 0.15 0.10 0.05 0.00 0 2 4 Succe ss 6 Sca t t e r pl ot of t hr e e f our t hsv sSucce ss 0.30 t h r e e f our t hs 0.25 0.20 0.15 0.10 0.05 0.00 0 2 4 Succe ss 6 8 10 Calculating Descriptive Statistics > Open the class survey results that were entered into the MINITAB worksheet. 2. Calculate descriptive statistics for the variable where students flipped a coin 10 times. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to the coin. The output will show up in your Session Window. Type the mean and the standard deviation here. Mean: 4.600 Standard deviation: 1.429 Short Answer Writing Assignment Both the calculated binomial probabilities and the descriptive statistics from the class database will be used to answer the following questions. 3. List the probability value for each possibility in the binomial experiment that was calculated in MINITAB with the probability of a success being . (Complete sentence not necessary) P(x=0) P(x=1) P(x=2) P(x=3) P(x=4) P(x=5) .000977 .009766 .043945 .117188 .205078 .246094 P(x=6) P(x=7) P(x=8) P(x=9) P(x=10) .205078 .117188 .043945 .009766 .000977 4. the Give probability for the following based on the MINITAB calculations with the probability of a success being . (Complete sentence not necessary) P(x1) .999023 P(x>1) .989258 P(4<x 7) .568359 P(x<0) 0 P(x4) .376953 P(x<4 or x7) .343750 5. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being . Either show work or explain how your answer was calculated. Mean = np, Standard Deviation = npq Mean: mean= np mean= (10)(.5) mean= 5 Standard deviation: SD= root(npq) SD=root (10)(.5)(.5) STDdev= root 2.5 STDdev= .581139 6. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being and compare to the results from question 5. Mean = np, Standard Deviation = npq Mean: Mean= np Mean= 10(.25) Mean= 2.5 Standard deviation: STDdev= root(npq) STDdev= root 10(.25)(.75) STDdev= root 1.875 STDdev= 1.369306 Comparison: In comparison to the results of question 5, or the probability of success being .5, the mean has a difference of 2.5 which makes sense since we are solving now for (half of .5). The standard deviation varies greatly though and seems to get higher the lower the chance of success, due to the skewed right graph of 1/4. The standard deviation in question 6 is . 788167 higher than the deviation of question 5. 7. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being and compare to the results from question 6. Mean = np, Standard Deviation = root npq Mean: Mean= np Mean= (10)(.75) Mean= 7.5 Standard deviation: STDdev= root(npq) STDdev= root 10(.75)(.25) STDdev= root 1.875 STDdev= 1.369306 Comparison: In comparison to the results from question 6, the mean has jumped from 2.5 to 7.5 due to the jump to chance of success. The part that is intriguing is the standard deviation is exactly the same. This is because if you look at the scatterplot for it is a mirror image of but now skewed left! It is perfectly symmetrical over x=5 between the 2 charts of success and success. 8. Explain why the coin variable from the class survey represents a binomial distribution. A binomial distribution can be anything being tested that has only 2 results. In this case we are flipping a coin. There are only 2 results that can occur from flipping the coin, heads or tails. That is all the possible outcomes, therefore making it a binomial distribution of statistics. 9. Give the mean and standard deviation for the coin variable and compare these to the mean and standard deviation for the binomial distribution that was calculated in question 5. Explain how they are related. Mean = np, Standard Deviation = root npq Mean: 4.600 Standard deviation: 1.429 Comparison: The mean and standard deviation of the coin variable is different than the results of question 5 because of several reasons. First, the mean is different since the binomial distribution assumes perfect statistical probability that you will flip 5 heads and 5 tails. The actual experiment yielded a mean of 4.6 heads/tails for 10 flips. The longer the experiment was to be repeated, the further it would become closer to the perfect mean of 5. Still 4.6 is within the binomial distributions standard deviation of 1.429, which is the same as the standard deviation of the coin flipping experiment.
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