Statistics – Lab #6
Name:______
Statistical Concepts:
· Data Simulation
· Discrete Probability Distribution
· Confidence Intervals
Calculations for a set of variables
Ø Open the class survey results that were entered into the MINITAB worksheet.
Ø 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.
Ø 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
Ø 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
Ø Open the class survey results that were entered into the MINITAB worksheet.
Ø 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.
Ø 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 (SEE THE OUTPUTS)
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.
It is an example of a discrete variable because the sample space is finite, S={1,2,3,4,5,6}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: ______1.708______Mean = E(x) = 1(1/6)+2(1/6)+….6(1/6) = (1+2+…+6)/6 = 21/6 = 3.5
Variance = V(x) = E(x2)-E2(x)
E(x2) = 12(1/6)+22(1/6)+….62(1/6) = (1+4+9+..+36)/6 = 91/6
V(x) = 91/6 –(7/2)2 = 91/6 – 49/4 = (182-147)/12 = 35/12
Standard deviation = √V(x) = √(35/12) = 1.708 (rounded to 3 dp)
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)?
Mean of the mean column is = 3.56, 3.56 it is very close to the parameter of interest (3.5) but is not equal to it4. 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)?
Mean of the median column is = 3.6, is close to the mean of 3.5 but is not as close as the answer from question 35. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
Sd of the mean = 0.106 and Sd of the median = 0.169, the first one is less than the second one so the variable “Mean” has less variability than the variable “Median”6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of interest? Explain your reasoning.
The mean is a better estimate because it is closer to the mean of 3.5 and the variability is lower7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
CI = (6.214,7.686) , We are 95% confident that the average of hours of sleep a students gets is between 6.214 and 7.6868. Give and interpret the 99% confidence interval for the hours of sleep a student gets.
CI = (5.944,7.956) , We are 99% confident that the average of hours of sleep a students gets is between 5.944 and 7.9569. 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.
Formula for CI is: x-bar ± t(a/2,n-1)s/√nThe 99% CI is wider than the 95% interval, both are centered at: x-bar =6.95 (mean of the mean) , the 99% CI is wider because the confidence level is greater, this produces a higher t-score so the CI is wider.