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Sample Mean Distribution And T-Interval

Sample Mean Distribution And T-Interval

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:

 

  • 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.

 

  • 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:

 

  • Apply the t-interval procedure to these data to find a 90% confidence 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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