Question

Ms. Gibbs’ 6th period chapter eight exam
Student Name	Exam Score
Robert Johnson	76
LaDavia Abrams	42
Jacobo Indiano	91
Johnathan Espy	95
Thomas Hartman	43
Stephanie Millerman	56
Alexis Abellan	46
William Press	85
John Mitchell	24
Brian Russell	25
Lisa Owens	60
Josue Alvarado	97
Alice Hsu	71
Raul Barrueco	99
Kathleen Drude	74
William Fritsche	90
Dwight Hare	72
Tommie Walker	92
Yolanda Pollard	68
Frank Bruscato	61
Larry Smith	85
Tonya Lumadue	53
Clifton Webb	81
John Cho	61
Sasha Gulley	72
Eishin Tanaka	77

As the school’s test coordinator/statistician, you are tasked with creating a report that includes the following results for Ms. Gibbs:

1. Create a Pareto chart (any other chart is incorrect! include all components of a Pareto chart to receive credit) using the following grading scale:

            A = 90 - 100

            B = 80 – 89

            C = 70 – 79

            D = 60 – 69

            F = 0 – 59

2. Calculate the Mean, Median, Mode, and Mid-range for the exam scores. I am sure that Ms. Gibbs understands the concept of “mean”, but you must to explain “mode” and “mid-range” to her in this report (failure to include explanations of mode and mid-range will result in a loss credit).

3a. Are the data positively skewed, symmetric, or negatively skewed (Yes or No)? Include the use a histogram in your explanation for part 3b.

3b. Include the histogram that you used to answer question 3a. Explain skewness to Ms. Gibbs in this report (failure to do so will result in a points being deducted).

4a. Calculate the Variance and Standard Deviation of the data.

4b. As the test coordinator, interpret each student’s test score - not the class as a group in terms of Standard Deviation. Ms. Gibbs never took statistics in her teacher education program.

5. Construct a boxplot of the test data.

6a. Create a table that lists the z – score for each student’s exam.

6b. As the test coordinator, interpret each students’ test score in terms of the calculated z-score.

7. Assess the data to determine if it is approximately normally distributed (Review the section in the book that discusses how to test for normality). You must include your calculations to support your answer (a graph is not a calculation). If you don’t include your calculations, you will not be given credit. Explain what normally distributed means to Ms. Gibbs.

8. Calculate the 95% Confidence Interval for the exam data. Once you do this, explain your calculated C.I. to Ms. Gibbs.

9. Ms. Gibbs is concerned about Kathleen Drude and John Cho. Since you are the data analyzer for the school, explain their exam performance relative to each other.

This assignment must be submitted on/before the due date. Please use the attached document to submit your responses.

Answer 1

1)

Add the Grades to each student on the basis of given range.

Steps to create Parento Chart:

1. Select the range A1:B10
2. On the Insert tab, in the Charts group, click the Histogram symbol
3. Click Pareto

Note this only works if you are using Excel 2016 or higher. If you are using older version, then sort the data in descending order, Calculate cumulative Frequency percentage. Now plot a combination chart and add the Frequency percentage to secondary axis.

2)

Mean = Total of Values/Count of Values = 1796/26 = 69.08

Median is the middle value when arranged in ascending order. Since the total count is even, Median will be the average of the two middle values.

Median = (72+72)/2 = 72

Mode is the most-common or most-frequently occurring value in a data set.

Since in the sample both 61 and 72 are repetiting twice, hence both are mode

Hence Mode = 61,72

Midrange is the average of maximum and minimum value. Hence,

Mid-Range = (24+99)/2 = 61.5

3)

a) The data is negatively skewed.

We create groups each of 10 unit’s width and calculate frequency for each group. Then we plot the histogram.

b)

A distribution is skewed if one of its tails is longer than the other. Since our histogram has a long tail in the negative direction it is negative skewed.

4)

a)

Standard deviation:

Using the above formulae,

Standard Deviation of the data(?) = 21.13

Variance = ? ² = 446.39

b)

Each Students Score can be interpreted in terms of standard deviation from mean.

Eg: For John Mitchell Score = 24 which is (Score – Mean)/Std Dev = (24-69.08)/21.13 = - 2.13 ie John’s Score is 2.13 Standard Deviation away from the class average.

Similarly, Raul Barrueco Score = 99 which is 1.42 ie Raul’s Score is 1.42 Standard Deviation away from the class average.