AP Statistics Practice Exam 2: answers and scoring guide

Example: "For adults who live in this city, what proportion use the library's app to borrow e-books?"

A good question names the group (adults in the city) and the variable (uses the app or not), and asks about all of them, not one person.

Treat each district as a stratum (a group).

In each district, number the adults on a list of residents and use a random number generator to pick 50 different adults.

The 4 × 50 = 200 adults chosen this way are the sample.

App use may be very different from district to district.

A stratified sample makes sure every district is in the sample, so the estimate for the whole city varies less from sample to sample.

This is a convenience sample: the adults are not chosen at random.

People who walk into the library already use the library more than other adults, so they are more likely to use its app.

So the estimate of the proportion who use the app would likely be too high.

The question

Question: find the mean of 18, 22, 25, 19, 30, 21, 24, 20 and 46.

The mean is the fair share.

We add them all, then share the total equally.

1 Add them all

We add all the numbers, from left to right.

So the total is 225.

2 Count them

Now we count the numbers.

There are 9.

3 Share equally

We share the total equally: 225 divided by 9.

225 divided by 9 is 25.

4 Check

Let's check: if every number were 25, 9 of them would make 25 times 9.

That is 225, the total.

So it is right!

Let's look back.

First: add them all.

Then: count them.

Then: share equally.

Last: check.

The question

Question: find the outliers in 18, 22, 25, 19, 30, 21, 24, 20 and 46.

An outlier is a number far away from the others.

1 Put them in order

First we put them in order, smallest first: 18, 19, 20, 21, 22, 24, 25, 30 and 46.

2 Find the quartiles

The median is the middle number.

9 numbers: 4 on each side of the middle one.

The median is 22.

The median cuts the list in two halves, 4 numbers each.

The median itself is in neither half.

Careful!

The median is not counted in the halves.

Each half has 4 numbers.

The quartiles cut the list into four quarters.

Q1 is the middle of the lower half.

Q3 is the middle of the upper half.

The lower half has two in the middle: 19 and 20.

Q1 is halfway between them: we add them and divide by 2.

19 plus 20 is 39.

Divided by 2 is 19.5.

The upper half has two in the middle: 25 and 30.

Q3 is halfway between them: we add them and divide by 2.

25 plus 30 is 55.

Divided by 2 is 27.5.

3 Find the IQR

The IQR, the interquartile range, is Q3 minus Q1: how wide the middle half is.

27.5 minus 19.5 is 8.

4 Find the fences

How far is far?

We need a rule.

The usual rule: more than 1.5 IQRs below Q1, or above Q3.

Everyone uses 1.5, so everyone finds the same outliers.

1.5 times 8 is 12.

A fence marks where far begins.

The low fence: Q1 minus 12.

19.5 minus 12 is 7.5.

The high fence: Q3 plus 12.

27.5 plus 12 is 39.5.

5 Find the outliers

On a number line: the fences in red, and a dot for each number.

46 is outside the fences.

So the outlier is 46.

6 Check

Let's check each one against its fence.

46 is more than 39.5.

So it is right!

Let's look back.

First: put them in order.

Then: find the quartiles.

Then: find the IQR.

Then: find the fences.

Then: find the outliers.

Last: check.

The mean (25 minutes) is larger than the median (22 minutes).

The long time of 46 minutes pulls the mean up, so the distribution is skewed to the right.

The median (22 minutes) for the typical time and the IQR (8 minutes) for the spread.

The data are skewed and have an outlier (46).

The median and the IQR are not pulled by the outlier; the mean and the standard deviation are.

A one-sample t-interval for a population mean.

μ = the mean weight of all bags of these crisps sold in the country.

Random: the bags are a random sample. 10%: 20 bags are less than 10% of all bags sold.

Normal: n = 20 is less than 30, but the dotplot shows no strong skew and no outliers, so a t-interval is reasonable.

The question

Question: a sample of 20 has mean 48.6 and SD 2.5: find the 95% confidence interval for the true mean.

x̄, the sample mean, is 48.6: the middle of our interval.

s is the SD of the sample: 2.5.

And n is 20.

1 Why t

We do not know σ, the SD of all the numbers.

We only have s, the SD of the sample.

So we use t instead of z.

The t curve is a bit wider than the bell, more so for small samples: a small sample makes us less sure.

It makes the interval a bit wider, to be safe.

The t-table has a row for each df, the degrees of freedom: n minus 1.

20 minus 1 is 19.

2 Find t*

95 percent of the bell is in the middle.

That leaves 5 percent: 2.5 percent in each end.

t* marks the cut between the middle and the ends.

In the t-table we take row 19.

Its columns are named by the middle part: the 95 percent column, which leaves 2.5 percent in each end.

Row 19, column 95%: t* = 2.093.

Careful!

df is n minus 1: 19, not 20.

3 Error bound

The error bound: t* times s over the square root of n.

s over the square root of n is the SD of sample means, from the sample means lesson.

t* of those SDs on each side holds the middle part we want.

2.5 over the square root of 20 is about 0.559.

Times 2.093: about 1.17.

4 The interval

Take the error bound away from the middle: 48.6 minus 1.17 is 47.43.

And add it: 48.6 plus 1.17 is 49.77.

The 95 percent confidence interval is (47.43, 49.77).

5 Say it in words

In words: we are 95 percent confident that the true mean is between 47.43 and 49.77.

6 Check

Let's check: the middle of the interval should be 48.6.

47.43 plus 49.77, divided by 2, is 48.6.

Half the width should be the error bound: 49.77 minus 47.43, divided by 2, is 1.17.

So it is right!

Let's look back.

First: why t.

Then: find t*.

Then: error bound.

Then: the interval.

Then: say it in words.

Last: check.

Yes. 50 grams is not in the interval (47.43, 49.77): every plausible value is below 50.

So there is convincing evidence that the mean weight of all bags of these crisps is less than 50 grams.

The question

Question: find the line of best fit for (20, 30), (22, 35), (24, 37), (26, 43), (28, 45).

The line of best fit is the straight line that comes closest to all the points.

1 Find the means

The mean of the x numbers: 120 divided by 5 is 24.

The mean of the y numbers: 190 divided by 5 is 38.

The line of best fit always goes through the mean point, (24, 38).

2 Make the table

A table with the points.

Each x minus the mean x, 24.

Each y minus the mean y, 38.

For the first point: 20 minus 24 is −4, and 30 minus 38 is −8.

The product: the two distances multiplied.

−4 times −8 is 32.

And the x distance squared.

−4 squared is 16.

We add the last two columns: 76 and 40.

3 Find the slope

A line is y = m x + b.

The slope m is how much y goes up when x goes up by 1.

The products show how y moves with x.

The squares show how x moves by itself.

So the slope is the total of the products, over the total of the squares.

Careful!

The bottom uses only the x distances, squared.

Not the y distances.

76/40 is 1.9.

4 Find b

b is where the line crosses the y axis.

The line goes through the mean point.

So the mean y is m times the mean x, plus b.

To get b alone, we take m times the mean x away from both sides.

1.9 times 24 is 45.6.

38 minus 45.6 is −7.6.

So the line of best fit is y = 1.9x − 7.6.

5 Draw the line

Now the points on a graph.

Each point is a dot.

The line goes through the mean point.

The line y = 1.9x − 7.6: it passes close to all the points.

6 Check

Let's check: the mean point must be on the line.

1.9 times 24 plus −7.6 is 38.

That is the mean y.

So it is right!

Let's look back.

First: find the means.

Then: make the table.

Then: find the slope.

Then: find b.

Then: draw the line.

Last: check.

For each extra advert in a week, the shop is predicted to get about 1.9 more new customers.

("Predicted": the line gives an average pattern, not the exact count in each week.)

Predicted: 1.9 × 24 − 7.6 = 45.6 − 7.6 = 38 new customers. Actual: 37.

Residual = actual − predicted = 37 − 38 = −1: that week the shop got 1 new customer fewer than the line predicts.

The question

Question: find the correlation coefficient r for (20, 30), (22, 35), (24, 37), (26, 43), (28, 45).

r is a number from negative 1 to 1.

It tells how close the points are to a straight line, going up or down.

1 Find the means

The mean of the x numbers: 120 divided by 5 is 24.

The mean of the y numbers: 190 divided by 5 is 38.

2 Make the table

A table with the points.

How far each point is from the means.

20 minus 24 is −4, and 30 minus 38 is −8.

These distances always add up to 0.

A good check.

The product: the two distances multiplied.

−4 times −8 is 32.

Why multiply?

Up and right of the means, both distances are positive: the product is positive.

Down and left, both are negative: positive again.

Those points make an uphill line.

The other points give negative products: downhill.

And each distance squared.

−4 squared is 16.

−8 squared is 64.

The squares show how spread out x and y are.

We add the last three columns: 76, 40 and 148.

3 Put them in the rule

The rule: the total of the products, over the square root of the two totals of squares multiplied.

Dividing by the spread keeps r between negative 1 and 1, whatever the numbers.

We put the totals in: 76 on top.

40 times 148 under the root.

Careful!

Do not forget the square root.

Without it, r comes out far too small.

4 Work it out

40 times 148 is 5920.

With a calculator, the square root of 5920 is about 76.942.

So r is about 76/76.942.

Divided: about 0.988, to 3 decimal places.

5 What r tells us

On a scale from negative 1 to 1.

r is here, at 0.988.

Positive: as x goes up, y goes up.

Close to 1: strong.

A strong positive correlation.

6 Check

Let's check.

The products add up to 76, so r must be positive.

It is.

And r is between negative 1 and 1.

So it makes sense!

Let's look back.

First: find the means.

Then: make the table.

Then: put them in the rule.

Then: work it out.

Then: what r tells us.

Last: check.