Module - Exercise
Title: Sample Mean Distribution and T-Interval
Task 1: Read the following case study, titled "Green M&M's":
Green M&M's: Consider a class of 20 statistics students, where each student is given 5 small bags of M&M's and asked to count the number of green M&M's. The results are shown below. The population mean is for a bag of this size is 10.06 and the population standard deviation is 2.59.
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Bag 1
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Bag 2
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Bag 3
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Bag 4
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Bag 5
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Student 1
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3
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4
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7
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6
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14
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Student 2
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12
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14
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18
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8
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3
|
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Student 3
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12
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18
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8
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13
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11
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Student 4
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18
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12
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7
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11
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8
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Student 5
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17
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18
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9
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14
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2
|
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Student 6
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3
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10
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14
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9
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13
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Student 7
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3
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12
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11
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9
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15
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Student 8
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6
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2
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3
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18
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11
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Student 9
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8
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16
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12
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17
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3
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Student 10
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14
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13
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11
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17
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5
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Student 11
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3
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14
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17
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17
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15
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Student 12
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7
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14
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11
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7
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2
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Student 13
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17
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2
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12
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18
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13
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Student 14
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9
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18
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8
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11
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10
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Student 15
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14
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16
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4
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3
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12
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Student 16
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4
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3
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7
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11
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14
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Student 17
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11
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17
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6
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5
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13
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Student 18
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15
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8
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17
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11
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10
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Student 19
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4
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9
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13
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16
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16
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Student 20
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12
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12
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5
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14
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16
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Answer the following questions:
- Find the sample means for each student's green M&M count.
- Create a histogram of the sample means, and calculate the mean and standard deviation of the sample means. To construct a histogram, follow the given steps:
- I. Obtain a frequency (relative-frequency, percent) distribution of the data.
- II. Draw a horizontal axis on which to place the bars and a vertical axis on which to display the frequencies (relative frequencies, percents).
- III. For each class, construct a vertical bar whose height equals the frequency (relative frequency, percent) of that class.
- IV. Label the bars with the classes, the horizontal axis with the name of the variable, and the vertical axis with "Frequency" ("Relative frequency," "Percent").
Source: Weiss, Neil A. (2012). Elementary Statistics (8th ed.). Upper Saddle River, NJ: Pearson.
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Sample Mean
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6.8
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11
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12.4
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11.2
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12
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9.8
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10
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8
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11.2
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12
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13.2
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8.2
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12.4
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11.2
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9.8
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7.8
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10.4
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12.2
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11.6
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11.8
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Class Interval
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Frequency
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Reative Frequencies
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6.05 - 7.05
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1
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0.067
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7.05 - 8.05
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2
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0.133
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8.05 - 9.05
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1
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0.067
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9.05 - 10.05
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3
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0.200
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10.05 - 11.05
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2
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0.133
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11.05 - 12.05
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7
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0.467
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12.05 - 13.05
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3
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0.200
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13.05 - 14.05
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1
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0.067
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Theoretically for all possible sample means for a sample size of 5, the mean = 10.06 , standard deviation = 2.59/√5= 1.1582832 and the distribution is normal.
Theoretically, what are the mean, standard deviation, and distribution of all possible sample means for a sample size of 5?
Theoretically for all possible sample means for a sample size of 5, themean = 10.06, standard deviation = 2.59/√5= 1.1582832 and the distribution is normal.
Task 2: Read the following case study, titled "Diamond Pricing":
In a Singapore Edition of Business Times, diamond pricing was explored. The price of a diamond is based on the diamond's weight, color, and clarity. A simple random sample of 18 one-half-carat diamonds had the following prices, in dollars:
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1676
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1442
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1995
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1718
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1826
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2071
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1947
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1983
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2146
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1995
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1876
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2032
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1988
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2071
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2234
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2108
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1941
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2316
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Based on the above information, solve the following problems:
Apply the t-interval procedure to these data to ?nd a 90% con?dence interval for the mean price of all one-half-carat diamonds. Interpret your result. (Note: ) Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data.
From the given data,
Sample mean = 1964.722222; Sample SD = 206.4526169 and as sample size is 18 so df = 18-1 = 17 so t-multiplier = t(0.05,17) = 2.110.
Thus the 90% confidence interval is,
90% CI = (1964.722222±2.110*206.4526169/√18) = (1862.047, 2067.398)
And the required plots are given below,
Stem-and-Leaf Display: Price
Stem-and-leaf of Price N = 18
Leaf Unit = 10
1 14 4
1 15
2 16 7
3 17 1
5 18 27
(6) 19 448899
7 20 377
4 21 04
2 22 3
1 23 1
b. Based on your graphs from part (b), is it reasonable to apply the t-interval procedure as you did in part a? Explain your answer.
The above graphs indicate presence of outlier and normality condition is violated due to that. Hence based on the graphs from part (b), it is not reasonable to apply the t-interval procedure.
Attachment:- Module.rar