Statistics: Concepts and Controversies
Statistics: Concepts and Controversies
9th Edition
ISBN: 9781464192937
Author: David S. Moore, William I. Notz
Publisher: W. H. Freeman
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Chapter 18, Problem 26E

(a)

To determine

To find: The mean of the scores of 10 students.

(a)

Expert Solution
Check Mark

Answer to Problem 26E

Solution: The required mean is 69.4.

Explanation of Solution

Calculation:

The mean of the exam scores of the 10 students is obtained as:

Mean=Sum of exam scoresTotal number of students=82+62+80++6210=69410=69.4

(b)

To determine

Section 1:

To find: A simple random sample of four students.

(b)

Expert Solution
Check Mark

Answer to Problem 26E

Solution: The selected students are those who are labeled as 1, 9, 2, and 3.

Explanation of Solution

Calculation:

The samples of four students are chosen using Table A of random digits provided in the book. The one digit random numbers are chosen from the table starting from the 101st row and the 1st column of the table. The random number is selected between 0 to 9. If the same random number is repeated after appearing for the first time, the number will be rejected. The selected random numbers are 1, 9, 2, and 3.

The students who are labeled with the same number of chosen random number are selected as sample.

Section 2:

To calculate: The sample mean of the scores of 4 randomly selected students.

Solution: The required mean is 65.5.

Explanation:

Calculation:

The scores of the selected students are 62, 62, 80, and 58. The sample mean (x¯) of the exam scores of the four randomly selected students is obtained as:

x¯=Sum of exam scores of selected studentsTotal number of students=62+62+80+584=2624=65.5

Interpretation: The mean value of the exam scores of 4 students is the estimate of population mean that is equal to 65.5.

(c)

To determine

Section 1:

To find: Ten simple random samples of four students.

(c)

Expert Solution
Check Mark

Answer to Problem 26E

Solution: The selected students are those who are labeled with the same number of selected random number. The below table shows the selected students:

Sample Selected Students
Sample 1 7, 3, 6, 4
Sample 2 4, 5, 6, 7
Sample 3 5, 2, 7, 1
Sample 4 9, 5, 2, 4
Sample 5 8, 2, 7, 3
Sample 6 6, 0, 9, 4
Sample 7 3, 6, 0, 9
Sample 8 3, 8, 4, 7
Sample 9 5, 9, 6, 3
Sample 10 6, 2, 5, 6

Explanation:

Calculation:

A sample of four students is chosen using Table A of random digits provided in the book. One-digit random numbers are chosen from the table starting from the 102th row and the 1st column of the table. A random number is selected between 0 and 9. If the same random number is repeated after appearing for the first time, it will be rejected. The selected random numbers are 7, 3, 6, and 4. So, the selected students for the first sample are the students who are labeled with the same number of chosen random numbers.

The above mentioned process is repeated for nine more times to obtained nine sample of size four using different row and column of Table A. The obtained result is shown in the below table:

Sample Random Number
Sample 1 7, 3, 6, 4
Sample 2 4, 5, 6, 7
Sample 3 5, 2, 7, 1
Sample 4 9, 5, 2, 4
Sample 5 8, 2, 7, 3
Sample 6 6, 0, 9, 4
Sample 7 3, 6, 0, 9
Sample 8 3, 8, 4, 7
Sample 9 5, 9, 6, 3
Sample 10 6, 2, 5, 6

The students who are labeled with the same number of chosen random numbers are selected as sample.

Section 2:

To find: The sample mean of the scores of 10 samples of size 4.

Solution: The required mean are shown in below table:

Sample Sample mean
Sample 1 65
Sample 2 69
Sample 3 70.25
Sample 4 71.75
Sample 5 69.5
Sample 6 70.25
Sample 7 66.75
Sample 8 67.5
Sample 9 64.5
Sample 10 70.75

Explanation of Solution

Calculation:

The scores of the students of first sample are 66, 58, 65, and 72. The sample mean (x¯1) of the exam scores of the first sample is obtained as:

x¯1=Sum of exam scores of selected studentsTotal number of students=66+58+65+724=2614=65.25

The calculation for the sample mean of the rest of the samples is shown in the below table:

Sample Selected Students Scores Sample Mean
Sample 2 4, 5, 6, 7 72, 73, 65, 66 x¯2=72+73+65+664=69
Sample 3 5, 2, 7, 1 73, 80, 66, 62 x¯3=73+80+66+624=70.25
Sample 4 9, 5, 2, 4 62, 73, 80, 72 x¯4=62+73+80+724=71.75
Sample 5 8, 2, 7, 3 74, 80, 66, 58 x¯5=74+80+66+584=69.5
Sample 6 6, 0, 9, 4 65, 82, 62, 72 x¯6=65+82+62+724=70.25
Sample 7 3, 6, 0, 9 58, 65, 82, 62 x¯7=58+65+82+624=66.75
Sample 8 3, 8, 4, 7 58, 74, 72, 66 x¯8=58+74+72+664=67.5
Sample 9 5, 9, 6, 3 73, 62, 65, 58 x¯9=73+62+65+584=64.5
Sample 10 6, 2, 5, 6 65, 80, 73, 65 x¯10=65+80+73+654=70.75

Section 3:

To graph: The histogram that shows the mean of the 10 samples.

Graph: To obtain the histogram, below steps are followed in the Minitab software.

Step 1: Enter the sample means in the Minitab worksheet.

Step 2: Go to “Graph” and select “Histogram.” Choose “Simple” and click OK.

Step 3: Enter the graph variables column and click OK.

Statistics: Concepts and Controversies, Chapter 18, Problem 26E

Interpretation: The obtained graph shows that maximum number of sample has mean value near 70.

To determine: Whether the population mean lies at the center of the histogram.

Solution: The population means lies at the center of the histogram.

Explanation:

The population mean is obtained as 69.4 in part (a). From the obtained histogram, it can be said that the population mean lies almost at the center of the histogram that is the center of the histogram close to the population mean.

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