1112 words - 5 pages

Statistics – Lab #6

Name: Hemanshu Patel

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 ...view middle of the document...

1. When rolling a die, is this an example of a discrete or continuous random variable? Explain your reasoning.

It's a discrete random variable, because the values of the variable is one of the 6 values in the set {1,2,3,4,5,6}. Since this set is finite, the random variable is discrete. Also we only have a 1/6 chances when rolling a number. |

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: 3.5 Standard deviation: 1.707Mean = 1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6) = 3.5Standard Deviation= (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) = sqrt(2.196666…) = 1.7078 |

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)?

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 MaximumMean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000Mean of means = 3.560 It is very close to the parameter of interest but is not equal to it. You could 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. |

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)?

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 MaximumMean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000Mean of median = 3.600, the mean here is close to the mean in question 2, but is not as close as the answer from question 3. |

5. Give the standard deviation for...

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