E. Culminating Data Management Investigation · Culminating Data Management Investigation

Analysing the Data I: Summaries, Graphs and the Normal Model

E1.4About 75 minutes

Learning goals

I can choose numerical summaries (mean, median, standard deviation, quartiles, IQR) that answer my hypothesis.
I can calculate and interpret these summaries for two groups and compare them in context.
I can draw side-by-side box plots and dot plots or histograms that make a comparison clear.
I can check for outliers with the 1.5 × IQR rule.
I can decide whether a normal model fits my data and use it to estimate a proportion, then compare the estimate with the actual data.

Words to know

numerical summary
A single number that describes a data set, such as the mean, median, standard deviation or IQR.

graphical summary
A graph that shows the shape, centre and spread of data, such as a histogram, dot plot or box plot.

five-number summary
Minimum, Q1, median, Q3 and maximum; the numbers drawn in a box plot.

side-by-side box plots
Box plots of two or more groups drawn on the same scale so that their centres and spreads can be compared directly.

fences
Q1 − 1.5 × IQR and Q3 + 1.5 × IQR. Values beyond the fences are flagged as possible outliers.

normal model
Using a normal distribution with a data set's mean and standard deviation to estimate proportions, when the data are roughly symmetric and mound-shaped.

Let the data speak

The data are clean and organized. Now comes the part the whole project was built for: analysis. The curriculum asks you to interpret, analyse and summarize your data, for example by generating and interpreting numerical and graphical statistical summaries, by recognizing and applying a probability distribution model, and by calculating expected values. This lesson does the one-variable analysis for Amira's Hypothesis 1; the next lesson does the two-variable analysis; the lesson after that applies probability models and expected value.

Every calculation here uses the example data set from Gathering, Cleaning and Organizing the Data (made up for teaching). The methods are the ones you learned in One-Variable Data Analysis.

Choose summaries that answer the question

Hypothesis 1 compares two groups, so Amira needs, for each group:

centre: the mean and the median (if they disagree a lot, the data are skewed or have outliers);
spread: the standard deviation and the IQR;
shape: a dot plot or histogram;
a comparison graph: side-by-side box plots on one scale;
the benchmark: the proportion meeting the 8-hour guideline.

The data are a sample, so use the sample standard deviation s (divide by n − 1). Quartiles use the method from One-Variable Data Analysis: split the ordered data into a lower and an upper half (leaving out the median if n is odd), then Q1 and Q3 are the medians of those halves.

Worked example

The model investigation, Stage 9: summaries for the job group

The 22 sleep values for students with jobs, in order, are:

6.25, 6.25, 6.5, 6.75, 7.0, 7.0, 7.0, 7.25, 7.25, 7.25, 7.5, 7.5, 7.5, 7.5, 7.75, 7.75, 8.0, 8.0, 8.0, 8.0, 8.75, 8.75

Find the mean, median, quartiles, IQR, standard deviation, and check for outliers. (Σx = 163.5 and Σx² = 1224.875.)

Mean: x̄ = 163.5 ÷ 22 ≈ 7.432 h.
Median: n = 22 is even, so average the 11th and 12th values: (7.5 + 7.5) ÷ 2 = 7.5 h.
Quartiles: the lower half is the first 11 values; its median (6th value) is Q1 = 7.0. The upper half is the last 11 values; its median (the 17th value overall) is Q3 = 8.0. IQR = 8.0 − 7.0 = 1.0 h.
Standard deviation: Σ(x − x̄)² = Σx² − (Σx)² ÷ n = 1224.875 − 163.5² ÷ 22 = 1224.875 − 1215.102... ≈ 9.7727. Then s² = 9.7727 ÷ 21 ≈ 0.46537, so s ≈ 0.682 h.
Outliers: fences are 7.0 − 1.5(1.0) = 5.5 and 8.0 + 1.5(1.0) = 9.5. All values lie between 6.25 and 8.75, so there are no outliers.
Interpretation: a typical student with a job sleeps about 7.5 hours on a school night, and most are within about 0.7 h of the mean.

Job group: x̄ ≈ 7.43 h, median 7.5 h, Q1 = 7.0, Q3 = 8.0, IQR = 1.0 h, s ≈ 0.68 h, no outliers.

Worked example

The model investigation, Stage 10: summaries for the no-job group

The 18 sleep values for students without jobs, in order, are:

7.5, 7.75, 8.0, 8.0, 8.25, 8.25, 8.25, 8.5, 8.5, 8.5, 8.75, 8.75, 8.75, 9.0, 9.0, 9.25, 9.5, 9.75

Find the same summaries. (Σx = 154.25 and Σx² = 1327.8125.)

Mean: 154.25 ÷ 18 ≈ 8.569 h.
Median: average of the 9th and 10th values: (8.5 + 8.5) ÷ 2 = 8.5 h.
Quartiles: lower half = first 9 values, median (5th) = Q1 = 8.25; upper half = last 9 values, median = Q3 = 9.0; IQR = 0.75 h.
Standard deviation: 1327.8125 − 154.25² ÷ 18 = 1327.8125 − 1321.8368... ≈ 5.9757; s² ≈ 5.9757 ÷ 17 ≈ 0.35151; s ≈ 0.593 h.
Outliers: fences 8.25 − 1.125 = 7.125 and 9.0 + 1.125 = 10.125; none.
Compare: the no-job mean is higher by 8.569 − 7.432 ≈ 1.14 h (about 68 minutes), and the median is higher by 1.0 h. Spreads are similar (s ≈ 0.68 vs 0.59), so the difference in centre is large compared with the variation within each group.

No-job group: x̄ ≈ 8.57 h, median 8.5 h, Q1 = 8.25, Q3 = 9.0, IQR = 0.75 h, s ≈ 0.59 h. Students without jobs slept about 1.1 hours more on average.

Side-by-side box plots of the example data. The middle half of the job group (7.0 to 8.0 h) lies entirely below the middle half of the no-job group (8.25 to 9.0 h).

Dot plots of the same data, one dot per student, with each mean marked by a dashed line. Both groups are roughly symmetric and mound-shaped, with no outliers, which makes a normal model reasonable.

Applying a probability distribution model

When a group's data are roughly symmetric and mound-shaped with no outliers, a normal model N(x̄, s) can estimate proportions, as in Continuous Distributions and the Normal Distribution. A good investigation then checks the model against the actual data. If the model and the data agree, the model can be used for estimates (for example, for the whole population); if not, report that and rely on the data.

One detail: sleep was recorded to the nearest quarter hour, so a recorded value of 8.0 stands for real sleep from 7.875 to 8.125 hours. "Recorded 8 or more" therefore means "actual sleep at least 7.875". This is the same idea as the continuity correction you used for the normal approximation to the binomial.

Worked example

The model investigation, Stage 11: normal model check

Use a normal model for each group to estimate the proportion who meet the 8-hour guideline, and compare with the actual data (6 of 22 with jobs and 16 of 18 without jobs recorded 8 hours or more).

Job group, N(7.432, 0.682): z = (7.875 − 7.432) ÷ 0.682 ≈ 0.65. From the z-table, P(Z < 0.65) ≈ 0.7422, so P(meets) ≈ 1 − 0.7422 = 0.258.
Actual: 6 ÷ 22 ≈ 0.273. The model is close (within about 1.5 percentage points, about one-third of a student).
No-job group, N(8.569, 0.593): z = (7.875 − 8.569) ÷ 0.593 ≈ −1.17. P(Z < −1.17) ≈ 0.1210, so P(meets) ≈ 1 − 0.1210 = 0.879.
Actual: 16 ÷ 18 ≈ 0.889. Again very close.
Conclusion: the normal model describes both groups well. In the job group, only about one in four students meets the guideline; in the no-job group, almost nine in ten do.

Model estimates 25.8% (job) and 87.9% (no job); actual 27.3% and 88.9%. The normal model fits well and supports Hypothesis 1.

Report summaries the way a reader can use them

Round final summaries sensibly (data to the nearest 0.25 h justify about two decimal places), give units, and always say which group and what n each summary describes. Put a summary table in the report: rows = groups, columns = n, mean, median, s, Q1, Q3, IQR. Then interpret in words: "Students without jobs slept about 1.1 hours more on a typical school night."

Just for fun

The job group's box plot and the no-job group's box plot were introduced at a party. "We should hang out," said one. "We barely overlap," said the other.

Hands-on activity

Your investigation, step 6: one-variable analysis

You need: your clean data, a calculator, a spreadsheet (optional), graph paper or graphing software.

For your main numerical variable, calculate the mean, median, standard deviation (s), Q1, Q3 and IQR for each group (or for the whole data set if you have only one group). Do one group by hand and check it with a spreadsheet (=AVERAGE, =MEDIAN, =STDEV.S; note that spreadsheet quartile functions may use a slightly different method).
Check for outliers with the fences.
Draw side-by-side box plots on one scale, and a histogram or dot plot for each group.
Describe shape, centre and spread in two or three sentences for each group, then compare.
If the shape is roughly normal, use a normal model to estimate one proportion and compare it with the actual proportion.
Make a summary table for your report.

Practice first with the example data: use the 40 rows in the data table to check that you get the same numbers as Amira.

Your turn

Using the job group's summaries, find the range and compare it with the IQR.

Find the z-score of a student with a job who sleeps 6.25 hours.

Using the dot plots, describe the shape of the no-job distribution.

Why does Amira use s (dividing by n − 1) rather than σ?

The difference in means is about 1.14 h. Convert it to minutes.

In the box plots, what fraction of the job group sleeps less than 7.0 hours, approximately? Check with the data.

More practice: warm-up

What is the median sleep of the no-job group?

What are the fences for a group with Q1 = 8.25 and Q3 = 9.0?

Which group has the larger standard deviation?

Name two graphs suitable for comparing a numerical variable across two groups.

Why does a recorded value of 8.0 hours correspond to 7.875 to 8.125 actual hours?

More practice: on your own

Find the mean and median of all 40 sleep values (Σ = 317.75) and explain why they are close.

Using the normal model for the job group, estimate the proportion who sleep less than 7 hours (recorded), and compare with the data.

If Amira had mistakenly entered one no-job value of 8.5 as 85, how would the mean and median of that group change?

Write a two-sentence comparison of the two groups suitable for a report.

Explain why the box plots alone could not show that the job group has more students than the no-job group.

Challenge

Using the normal model for the no-job group, find the sleep time exceeded by only 10% of these students, and interpret it.

Amira's standard deviations are about 0.6 to 0.7 h, and the difference in means is about 1.14 h. Express the difference in units of the job group's standard deviation and explain why this helps judge whether the difference matters.

Suppose the job group's distribution had been strongly skewed left (a few students sleeping 3 to 4 hours). How should Amira's choice of summaries and model change?

Check your understanding

Quiz

In the example data, the median sleep of the job group is:

For a data set with Q1 = 7.0 and Q3 = 8.0, the upper fence is:

Which graph best compares the centres and spreads of two groups on one scale?

Why is the sample standard deviation (dividing by n − 1) used for Amira's data?

A normal model for the job group, N(7.432, 0.682), estimates about 25.8% meet the guideline; the data show 27.3%. This suggests:

Students without jobs slept about how much more, on average, than students with jobs?

Which statement is correct for comparing two groups' box plots?

For parents and teachers

Big idea (E1.4): good analysis chooses summaries and graphs that answer the hypothesis, compares groups fairly on the same scale, and checks any probability model against the actual data.

How to help: Ask your teen to explain the box plot in plain words ("the middle half of students with jobs sleep between 7 and 8 hours"). Check one calculation together, such as a median or a mean, using the data table. Ask, "Does the model agree with the real data? How do you know?"

Common difficulties:

Calculating everything but interpreting nothing: each number needs a sentence in context.
Mixing up σ and s, or using spreadsheet quartiles without noticing they use a different method.
Drawing box plots on different scales, which makes comparison misleading.
Applying a normal model to skewed data without checking the shape first.