12.03a - Estimating from grouped data

12.03a - Estimating from grouped data

Grouping makes a large data set easier to read, but hides the exact observations. You will find a modal class and estimate mean, median and range from grouped tables. You will make the assumptions behind each estimate clear, so an approximate answer is never mistaken for an exact one.

What grouped data tells us

Ten people finish a task, but their times are recorded only as “under 10 minutes”, “10 to under 20” and “20 to under 30”. Can we still work out their average time?

A class interval collects values within stated boundaries. In 10≤t<2010\le t<20, 10 is included and 20 is excluded; 20 belongs to the next class. Intervals should cover the intended values without overlapping or leaving unintended gaps. A frequency tells us how many observations lie inside its interval, not where they lie within it.

Modal class

The modal class is a class with the greatest frequency. If greatest frequencies tie, report the tied classes. Its interval is known, but the exact most frequent individual value usually is not.

For equal-width classes this identifies the busiest part of the scale directly. The tables in this part use equal widths. With unequal widths, a class may contain more observations partly because it covers more values; this matters when interpreting the shape of a distribution.

Worked example 1: the most frequent interval

Question. Identify the modal class and explain whether the exact times are known.

Time tt, minutesFrequency
0≤t<100\le t<103
10≤t<2010\le t<205
20≤t<3020\le t<302

Plan. Locate the largest frequency, then distinguish the class from individual values.

max⁡(3,5,2)=5greatest frequencymodal class:10≤t<20the interval attached to 5\begin{aligned} \max(3,5,2)&=5 &&\text{greatest frequency}\\ \text{modal class}&:10\le t<20 &&\text{the interval attached to 5} \end{aligned}

Answer. The modal class is 10≤t<2010\le t<20 minutes. The individual times are unknown; five different times, or repeated equal times, could fit that class.

Check. Values 11, 12, 13, 14, 15 and values 16, 16, 17, 18, 19 would each give a frequency of five in this class. The table cannot distinguish them.

Change only the last frequency from 2 to 5.

Worked example 2: a tied modal class

Question. The frequencies for the same three intervals are now 3, 5 and 5. State the modal class or classes.

Plan. Keep every interval whose frequency equals the greatest one.

greatest frequency=5matching intervals:10≤t<20,20≤t<30\begin{aligned} \text{greatest frequency}&=5 \\ \text{matching intervals}&:10\le t<20,\quad20\le t<30 \end{aligned}

Answer. Both 10≤t<2010\le t<20 and 20≤t<3020\le t<30 minutes are modal classes.

Check. Each contains five observations and the remaining interval contains only three. Selecting one tied class arbitrarily would discard information.

The boundaries 10 and 20 are not the “modes”. They label a class; they do not tell us which individual time occurred most often.

Key idea

Grouping keeps frequencies but loses individual values. Most numerical summaries calculated from grouped data therefore need estimates or assumptions.

Check which information survives grouping.

Estimating the mean with midpoints

Use the centre of each interval as a representative value when the individual values are hidden.

The midpoint is halfway between the class boundaries. In 10≤t<2010\le t<20 it is (10+20)/2=15(10+20)/2=15. Replacing every observation in that class by 15 gives an estimated total 15×f15\times f. It is a sensible central choice when we have no evidence that values cluster towards one end; it is not a claim that every observation equals 15.

estimated mean=sum of (class midpoint×frequency)total frequency\text{estimated mean}=\frac{\text{sum of }(\text{class midpoint}\times\text{frequency})}{\text{total frequency}}

The answer is an estimate because the actual class totals are unknown. If an interval has no finite upper or lower boundary, its midpoint cannot be calculated without more information.

Keep an extra product column so the weighting is visible.

Worked example 3: estimating a hidden total

Question. Estimate the mean time for the ten people in this table.

Time tt, minutesFrequency ffMidpointProduct
0≤t<100\le t<103515
10≤t<2010\le t<2051575
20≤t<3020\le t<3022550

Plan. Add midpoint products and divide by the number of people.

estimated total=15+75+50=140minutesn=3+5+2=10peopleestimated mean=140/10=14minutes per person\begin{aligned} \text{estimated total}&=15+75+50=140 &&\text{minutes}\\ n&=3+5+2=10 &&\text{people}\\ \text{estimated mean}&=140/10=14 &&\text{minutes per person} \end{aligned}

Answer. An estimated mean of 14 minutes.

Check. The midpoint data would be three 5s, five 15s and two 25s. Their mean must be below 15 because there is one more value at 5 than at 25. A mean of 14 fits that balance.

Change only the last frequency, from 2 to 4. More weight now lies in the upper interval.

Worked example 4: changing the weights

Question. Frequencies for the same intervals are now 3, 5 and 4. Estimate the mean, to one decimal place.

Plan. Retain the midpoints but recalculate the weighted total and count.

estimated total=5(3)+15(5)+25(4)=190n=3+5+4=12estimated mean=190/12≈15.8round at the end\begin{aligned} \text{estimated total}&=5(3)+15(5)+25(4)=190 \\ n&=3+5+4=12 \\ \text{estimated mean}&=190/12\approx15.8 &&\text{round at the end} \end{aligned}

Answer. Approximately 15.8 minutes.

Check. Two added midpoint values of 25 increase the old total by 50: 140+50=190140+50=190. The mean should rise above 14 but stay below 25, as it does.

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