Sample mean distribution and


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.
Bag 1 Bag 2 Bag 3 Bag 4 Bag 5
Student 1 3 4 7 6 14
Student 2 12 14 18 8 3
Student 3 12 18 8 13 11
Student 4 18 12 7 11 8
Student 5 17 18 9 14 2
Student 6 3 10 14 9 13
Student 7 3 12 11 9 15
Student 8 6 2 3 18 11
Student 9 8 16 12 17 3
Student 10 14 13 11 17 5
Student 11 3 14 17 17 15
Student 12 7 14 11 7 2
Student 13 17 2 12 18 13
Student 14 9 18 8 11 10
Student 15 14 16 4 3 12
Student 16 4 3 7 11 14
Student 17 11 17 6 5 13
Student 18 15 8 17 11 10
Student 19 4 9 13 16 16
Student 20 12 12 5 14 16
Answer the following questions:
a. Find the sample means for each student's green M&M count.
b. 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.
c. Theoretically, what are the mean, standard deviation, and distribution of all possible sample means for a sample size of 5?

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:
1676 1442 1995 1718 1826 2071 1947 1983 2146
1995 1876 2032 1988 2071 2234 2108 1941 2316

Based on the above information, solve the following problems:
a. 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.
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.

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Basic Statistics: Sample mean distribution and
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