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Statistics – Lab Week 4 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 ¼. 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 ½. 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 ¾. 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. Scatterplot of One fourth vs Sucess 0.30 0.25 0.20 One fourth 0.15 0.10 0.05 0.00 0 2 4 6 8 10 Sucess Scatterplot of One half vs Sucess 0.25 0.20 0.15 One half 0.10 0.05 0.00 0 2 4 6 8 10 Sucess Scatterplot of Three fourths vs Sucess 0.30 0.25 0.20 Three fourths 0.15 0.10 0.05 0.00 0 2 4 6 8 10 Sucess 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: 5.00 Standard deviation: 3.32 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 ½. (Complete sentence not necessary) P(x=0) 0.0009766 P(x=6) 0.205078 P(x=1) 0.0097656 P(x=7) 0.117188 P(x=2) 0.0439453 P(x=8) 0.0439453 P(x=3) 0.117188 P(x=9) 0.0097656 P(x=4) 0.205078 P(x=10) 0.0009766 P(x=5) 0.246094 4. Give the probability for the following based on the MINITAB calculations with the probability of a success being ½. (Complete sentence not necessary) P(x≥1) 0.99902 P(x<0) P(x>1) 0.0097656 P(x≤4) 0.376953 P(4<x ≤7) 0.568359 P(x<4 or x≥7) 0.2265625 5. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being ½. Either show work or explain how your answer was calculated. Mean = np, Standard Deviation = npq Mean: np = 10 * ½ = 5 1 1 Standard deviation: np 1 p 10* 1 = 1.58113883 2 2 6. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being ¼ and compare to the results from question 5. Mean = np, Standard Deviation = npq Mean: np = 10 * ¼ = 2.5 1 1 Standard deviation: np 1 p 10* 1 = 1.369306394 4 4 Comparison: Mean has been reduced to the half of that in question 5. Standard deviation is less than that in question 5. 7. Calculate the mean and standard deviation (by hand) for the MINITAB created binomial distribution with the probability of a success being ¾ and compare to the results from question 6. Mean = np, Standard Deviation = npq Mean = np = 10 * ¾ = 7.5 3 3 Standard deviation = np 1 p 10* 1 = 1.369306394 4 4 8. Explain why the coin variable from the class survey represents a binomial distribution. 1. There are a fixed number of trials of 10. That is, there are n trials. 2. Each trial results in either head or tail. That is, each trial has two possible outcomes, “success” or “failure”. 3. The probability of success is the same for each trial. 4. Each trial is independent. That is, the experiment satisfies all the conditions of a binomial experiment. Hence it is a binomial experiment. 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 = npq Mean & standard deviation for the coin variable Mean: 5.00 Standard deviation: 3.32 Mean & standard deviation for the binomial distribution Mean = np = 10 * ½ = 5 1 1 Standard deviation = np 1 p 10* 1 = 1.58113883 2 2 Mean is identical in both cases. . Statistics – Lab #5 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 next column in the Worksheet with the word median. Repeat the above steps but select the radio-button that corresponds to Median and in the Store results in: text area, place the median column. Calculating Descriptive Statistics Calculate descriptive statistics for the mean and median columns that where created above. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to mean and median. The output will show up in your Session Window. Print this information. Calculating Confidence Intervals for one Variable Open the class survey results that were entered into the MINITAB worksheet. We are interested in calculating a 95% confidence interval for the hours of sleep a student gets. Pull up Stat > Basic Statistics > 1-Sample t and set Samples in columns: to Sleep. Click the OK button and the results will appear in your Session Window. We are also interested in the same analysis with a 99% confidence interval. Use the same steps except select the Options button and change the Confidence level: to 99. Short Answer Writing Assignment All answers should be complete sentences. 1. When rolling a die, is this an example of a discrete or continuous random variable? Explain your reasoning. 2. Calculate the mean and standard deviation of the probability distribution created by rolling a die. Either show work or explain how your answer was calculated. Mean: ____________ Standard deviation: _________________ 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)? 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)? 5. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability? 6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of interest? Explain your reasoning. 7. Give and interpret the 95% confidence interval for the hours of sleep a student gets. 8. Give and interpret the 99% confidence interval for the hours of sleep a student gets. 9. Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain the difference between these intervals and why this difference occurs.

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posted: | 8/10/2012 |

language: | English |

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