End-of-topic test: Statistics

Unit 3 · Topic 3.2b end-of-topic test

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback tells you what a wrong choice assumed. For the two free-response questions, write your answer in full sentences and show any calculation, then open the scoring guide and mark your own work against it. Use the mean, the standard deviation and the standard error of the mean exactly as the AP formula sheet writes them; the standard deviation divides by n − 1, and an error bar of ±2SE covers the range the true mean is likely to lie in. Rounding: keep the full value in your calculator and write at least three significant figures in your working; report a mean to one more decimal place than the readings, and a standard deviation or standard error to three significant figures. Where a graph's legend says its error bars represent ±2SE, the overlap rule applies.

Question 1

Five tubes of catalase at 30 °C released oxygen at 4.2, 3.8, 4.5, 4.1 and 3.9 mL/min.

What mean rate should be reported for 30 °C?

A. 0.7 mL/min B. 4.10 mL/min C. 5.13 mL/min D. 20.5 mL/min

Question 2

Five trials with enzyme A gave rates with a mean of 8.0 mL/min and a standard deviation of 0.3 mL/min. Five trials with enzyme B gave a mean of 5.0 mL/min and a standard deviation of 0.9 mL/min.

Which statement do these values support?

A. Enzyme B's rate is higher, because its standard deviation is larger B. Enzyme A's five readings all lay within 0.3 mL/min of one another C. Enzyme B's readings were less accurate, so its mean of 5.0 should be dropped D. Enzyme B's readings varied more from one trial to the next

Question 3

Five readings of an enzyme's rate were 6.0, 6.4, 5.6, 6.2 and 5.8 μmol/min. Their mean is 6.00 μmol/min. Use the formula on the AP sheet, $s = \sqrt{\frac{\sum {(x_{i} - 𝑥̄)}^{2}}{n - 1}}$ .

What is the standard deviation of these readings?

A. 0.10 μmol/min B. 0.283 μmol/min C. 0.316 μmol/min D. 0.40 μmol/min

Question 4

Nine trials of a reaction gave a mean rate of 12.0 mL/min with a standard deviation of 0.60 mL/min.

What is the standard error of this mean?

A. 0.067 mL/min B. 0.20 mL/min C. 0.60 mL/min D. 1.80 mL/min

Question 5

The student's graph: mean rate of oxygen release by catalase at two temperatures, five trials each.

A student plots the mean rate of oxygen release by catalase at two temperatures, five trials each, and draws an error bar on each mean. Her legend on the graph reads "error bars = ±1 standard error". She concludes that catalase is faster at 30 °C than at 20 °C, because there is a clear gap between the two bars.

Which statement about her conclusion is correct?

A. Her bars are ±1SE; the overlap test needs ±2SE bars, which are twice as long and may overlap B. Her bars show where each true mean is likely to lie, so the conclusion is sound C. A gap between ±1SE bars is the overlap test itself, so no further check is needed D. Five trials is too few for any conclusion, whatever the bars show

Question 6

Mean rate of an enzyme at three pH values, five trials each, same temperature and substrate concentration. Error bars represent ±2SE.

The graph shows the mean rate of an enzyme at pH 6, 7 and 8.

What do the error bars show about the rates at pH 7 and pH 8?

A. The rates at pH 7 and pH 8 are the same B. These data do not show a difference between pH 7 and pH 8 C. pH 8 is faster, because its mean is higher D. Every reading at pH 8 lay above every reading at pH 7

Question 7

A student measured the length of six bean roots after three days of growth, from the seed to the root tip, with a ruler. The roots measured 34, 41, 38, 29, 45 and 37 mm.

What mean root length should be reported?

A. 16 mm B. 37.3 mm C. 44.8 mm D. 224 mm

Question 8

Five tubes of a protease at 35 °C released amino acids at 5.1, 5.8, 4.6, 5.3 and 4.7 mg/min. Their mean is 5.10 mg/min. Use the formula on the AP sheet, $s = \sqrt{\frac{\sum {(x_{i} - 𝑥̄)}^{2}}{n - 1}}$ .

What is the standard deviation of these readings?

A. 0.235 mg/min B. 0.434 mg/min C. 0.485 mg/min D. 0.94 mg/min

Question 9

Six tomatoes from one plant were weighed on a kitchen balance. Their masses have a mean of 86.5 g and a standard deviation of 1.20 g.

What is the standard error of this mean?

A. 0.200 g B. 0.490 g C. 0.537 g D. 2.94 g

Question 10

Two students each measured the same six sunflower seedlings with a ruler. Student 1's six readings have a mean of 41.5 mm and a standard deviation of 1.2 mm. Student 2's have a mean of 41.8 mm and a standard deviation of 4.9 mm.

Whose six readings agreed most closely with one another, and how can you tell?

A. Student 2's, because the mean of her readings is the larger B. Student 2's, because the standard deviation of her readings is the larger C. Student 1's, because the mean of her readings is the smaller D. Student 1's, because the standard deviation of her readings is the smaller

Question 11

A report on the mass of 25 bean seeds from one plant gives a mean of 0.412 g, a standard deviation of 0.035 g and a standard error of 0.0070 g.

Which statement about these values is correct?

A. The 0.0070 g says how far the sample mean is likely to sit from the true mean of all such seeds B. The 0.0070 g says how far a single seed's mass typically sits from the mean C. The 0.035 g says how far the sample mean is likely to sit from the true mean D. The 0.035 g is the mistake the balance made in weighing each seed

Question 12

A class measured the rate of a reaction in nine tubes and calculated the standard error of the mean. They repeat the experiment with thirty-six tubes, and the standard deviation of the readings comes out the same as before.

What happens to the standard error of the mean?

A. The standard error falls to a quarter of its value B. The standard error stays the same C. The standard error halves D. The standard error doubles

Question 13

Five tubes of a lipase at 30 °C released fatty acids at a mean rate of 8.40 μmol/min. The standard deviation of the five readings is 0.47 μmol/min and the standard error of the mean is 0.21 μmol/min.

Between which values does the ±2SE error bar on this mean run?

A. 7.46 to 9.34 μmol/min B. 8.19 to 8.61 μmol/min C. 8.40 to 8.82 μmol/min D. 7.98 to 8.82 μmol/min

Question 14

Three drawings of the same mean, 4.20 mL/min, each with an error bar meant to represent ±2SE. Gridlines every 0.1 mL/min.

A student measured a protease's rate at 20 °C in five tubes: mean 4.20 mL/min, standard error 0.15 mL/min. She drew the mean three times, each time with an error bar meant to represent ±2SE, as shown below.

Which drawing shows the ±2SE error bar correctly?

A. Drawing 1 B. Drawing 2 C. Drawing 3 D. None of the three drawings

Question 15

Mean rate of maltose release by an amylase at pH 6 and pH 7, five tubes each, same temperature and starch concentration. Error bars represent ±2SE. Gridlines every 0.2 mg/min.

The graph shows the mean rate of maltose release by an amylase at pH 6 and at pH 7, five tubes each, with error bars.

Between which values does the pH 7 error bar run, and what does it show?

A. 2.80 to 3.60 mg/min; the range the true mean is likely to lie in B. 4.20 to 4.60 mg/min; the range the true mean is likely to lie in C. 4.20 to 5.00 mg/min; the range the true mean is likely to lie in D. 4.20 to 5.00 mg/min; the highest and lowest of the five readings

Question 16

Mean rate of oxygen release by catalase at four alcohol concentrations, five tubes each, same temperature and pH. Error bars represent ±2SE. Gridlines every 0.5 mL/min.

The graph shows the mean rate of oxygen release by catalase at four alcohol concentrations, five tubes each, with temperature and pH the same in every tube.

Which pair of alcohol concentrations do these data show to have different rates?

A. 2% and 4% B. 0% and 2% C. 4% and 6% D. All three of these pairs, because each pair's means differ

Question 17

Five tubes of a protease at 20 °C released amino acids at a mean rate of 2.40 mg/min, with a ±2SE error bar from 2.10 to 2.70 mg/min. Five tubes at 25 °C released them at a mean rate of 2.60 mg/min, with a ±2SE error bar from 2.35 to 2.85 mg/min.

What may be concluded about the rates at 20 °C and 25 °C?

A. The rate at 25 °C is higher, because its mean is 0.20 mg/min above the other B. The two rates are the same, because the two bars overlap C. The rate at 25 °C is higher, because its bar reaches higher than the other D. These data do not show a difference between the two temperatures

Question 18

Mean rate of oxygen release by a catalase from spinach leaves at 30 °C and 60 °C, five tubes each, same pH and hydrogen peroxide concentration. Error bars represent ±2SE. Gridlines every 0.5 mL/min.

The graph shows the mean rate of oxygen release by a catalase from spinach leaves at 30 °C and at 60 °C, five tubes each, with error bars.

Which statement is supported by the graph and by what heat does to an enzyme?

A. The bars overlap, so these data do not show that the change in temperature affected the rate B. The bars are apart; at 60 °C the molecules moved too slowly to collide with the substrate often enough C. The bars are apart; at 60 °C the weak interactions holding the fold were disrupted, so the active site lost its shape D. The bars are apart; at 60 °C the extra collisions used the substrate up before the oxygen could be collected

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Analyze Data · 4 points

Students measured the rate at which an amylase extracted from germinating wheat seeds breaks starch down into the sugar maltose, at four temperatures, with five trials at each temperature. The pH and the starch concentration were the same in every trial. The table gives the mean rate at each temperature and the standard error of that mean (n = 5). The students plan to plot the four means as bars with error bars representing ±2SE.

Mean rate of maltose release by wheat amylase at four temperatures, five trials each. The standard error of each mean is given; n = 5.

(a) Describe the relationship between temperature and the mean rate shown in the table, using values from the table. (1 point)

A full-credit answer: The mean rate rises with temperature from 20 to 40 °C, from 2.20 to 4.40 to 7.20 mg of maltose per minute, and then levels off: at 50 °C the mean is 7.40 mg/min, close to the 40 °C value.

Check the box for each point your answer earns

the mean rate rises with temperature from 20 to 40 °C (2.20 to 4.40 to 7.20 mg/min) and is about the same from 40 to 50 °C (7.20 and 7.40 mg/min).

Accept "rises steeply, then levels off" with at least two values quoted. Do not award the point for a trend with no values, or for a description that has the rate falling between 40 and 50 °C.

Common slip: Writing "the rate increases with temperature" and stopping. A describe-the-data point needs the values and the whole pattern, including the leveling off from 40 to 50 °C.

(b) Calculate the range covered by a ±2SE error bar for 40 °C and for 50 °C. (1 point)

A full-credit answer: For 40 °C the bar runs from 6.94 to 7.46 mg/min, and for 50 °C from 7.01 to 7.79 mg/min.

40 °C: x̄ = 7.20 mg/min, SE = 0.131 mg/min

50 °C: x̄ = 7.40 mg/min, SE = 0.195 mg/min

bar ends = 𝑥̄ \pm 2 S E

bar ends = 𝑥̄ \pm 2 S E

40 °C: 2 S E = 2 \times 0.131 = 0.262 mg/min

40 °C: lower end = 7.20 - 0.262 = 6.94 mg/min

40 °C: upper end = 7.20 + 0.262 = 7.46 mg/min

50 °C: 2 S E = 2 \times 0.195 = 0.390 mg/min

50 °C: lower end = 7.40 - 0.390 = 7.01 mg/min

50 °C: upper end = 7.40 + 0.390 = 7.79 mg/min

Check the box for each point your answer earns

both ranges: 40 °C, 2SE = 0.262, so the bar runs from 6.94 to 7.46 mg/min; 50 °C, 2SE = 0.390, so the bar runs from 7.01 to 7.79 mg/min (accept each limit within ±0.01).

Accept the ranges written as 7.20 ± 0.26 and 7.40 ± 0.39. Do not award the point for bars of ±1SE, or for bars built from the standard deviation.

Common slip: Adding and subtracting one SE instead of two, or using the standard deviation. The bar the students planned is ±2SE, the range the true mean is likely to lie in.

(c) One student claims that the enzyme works faster at 50 °C than at 40 °C. Use the two error bars to evaluate the claim, and state whether the null hypothesis of no difference between 40 and 50 °C is rejected. (1 point)

A full-credit answer: The 40 °C bar (6.94 to 7.46) and the 50 °C bar (7.01 to 7.79) overlap, so the gap between the means, 7.40 against 7.20, could be chance. These data do not show a difference between the two temperatures, so the null hypothesis of no difference is not rejected. That is not the same as showing the two rates are equal.

Check the box for each point your answer earns

the two ±2SE bars overlap (6.94 to 7.46 shares its range with 7.01 to 7.79), so the difference between the means could be chance; the data do not show a difference, and the null hypothesis is not rejected.

Accept "the claim is not supported by these data". Do not award the point for "the rates are the same" (overlap shows no difference, not equality), or for a decision made from the two means alone.

Common slip: Reading overlapping bars as "the rates are the same". Overlap means the data have not shown a difference; the true means may still differ.

(d) The students propose repeating the experiment at 75 °C. Predict how the mean rate at 75 °C will compare with the mean at 50 °C, and justify your prediction in terms of the enzyme's structure. (1 point)

A full-credit answer: The mean rate at 75 °C will be far lower than at 50 °C, close to zero. Warming speeds collisions, but above the optimum the heat disrupts the hydrogen bonds and other weak interactions that hold the enzyme's fold, so the active site loses its shape, the starch no longer fits, and the denatured enzyme can no longer catalyze the reaction; that loss far outweighs the extra collisions.

Check the box for each point your answer earns

the mean rate at 75 °C will be much lower than at 50 °C (near zero), because above the enzyme's optimum heat disrupts the hydrogen bonds and other weak interactions that hold the protein's fold, so the active site loses its shape and the substrate no longer fits (denaturation), and this loss outweighs the extra collisions that warming brings.

Accept "denatured" only with what it does to the active site or to substrate binding. Do not award the point for "faster, because molecules move faster", or for "lower" with no structural reason.

Common slip: Predicting a higher rate because hotter molecules collide more often. That is true only below the optimum; the leveling off from 40 to 50 °C is already the sign that denaturation has begun to cancel the gain.

Free-response score: 0 of 4

Free response 2 · Analyze Data · 4 points

A lipase from a fungus is breaking a fat down, and the fatty acids released are measured each minute. Students ran five tubes at pH 8 and five at pH 6, with the same temperature and fat concentration in every tube. The five pH 8 tubes released fatty acids at 3.2, 4.0, 3.0, 3.6 and 3.2 mg/min. For the pH 6 tubes the students have already calculated a mean of 2.60 mg/min and a standard error of 0.120 mg/min, so its ±2SE error bar runs from 2.36 to 2.84 mg/min.

(a) Calculate the mean rate at pH 8. (1 point)

mg/min

Write down the values in the question:

x_{i} = 3.2, 4, 3, 3.6, 3.2 mg/min

n = 5

Write down the equation:

𝑥̄ = \frac{\sum x_{i}}{n}

Substitute the values into the equation:

𝑥̄ = \frac{\sum x_{i}}{n}

𝑥̄ = \frac{3.2 + 4 + 3 + 3.6 + 3.2}{5}

𝑥̄ = \frac{17}{5}

𝑥̄ = 3.40 mg/min

Round the value you report:

The readings have one decimal place, so the mean is reported to two: 3.40 mg/min.

A full-credit answer: The mean rate at pH 8 is 3.40 mg/min.

Accept 3.4 mg/min. Do not award the point for 17.0 (the sum) or for 4.25 (the sum divided by 4).

(b) Calculate the standard deviation of the pH 8 readings. (1 point)

mg/min

Write down the values in the question:

x_{i} = 3.2, 4, 3, 3.6, 3.2 mg/min

𝑥̄ = 3.40 mg/min

n = 5

Write down the equation:

s = \sqrt{\frac{\sum {(x_{i} - 𝑥̄)}^{2}}{n - 1}}

Subtract the mean from each reading and square the result:

{(x_{1} - 𝑥̄)}^{2} = {(3.2 - 3.40)}^{2} = {(−0.2)}^{2} = 0.04 {(mg/min)}^{2}

{(x_{2} - 𝑥̄)}^{2} = {(4 - 3.40)}^{2} = {(+0.6)}^{2} = 0.36 {(mg/min)}^{2}

{(x_{3} - 𝑥̄)}^{2} = {(3 - 3.40)}^{2} = {(−0.4)}^{2} = 0.16 {(mg/min)}^{2}

{(x_{4} - 𝑥̄)}^{2} = {(3.6 - 3.40)}^{2} = {(+0.2)}^{2} = 0.04 {(mg/min)}^{2}

{(x_{5} - 𝑥̄)}^{2} = {(3.2 - 3.40)}^{2} = {(−0.2)}^{2} = 0.04 {(mg/min)}^{2}

Add these values:

\sum {(x_{i} - 𝑥̄)}^{2} = 0.04 + 0.36 + 0.16 + 0.04 + 0.04

\sum {(x_{i} - 𝑥̄)}^{2} = 0.64 {(mg/min)}^{2}

Substitute the values into the equation:

s = \sqrt{\frac{\sum {(x_{i} - 𝑥̄)}^{2}}{n - 1}}

s = \sqrt{\frac{0.64}{5 - 1}}

s = \sqrt{\frac{0.64}{4}}

s = \sqrt{0.16}

s = 0.400 mg/min

Round the value you report:

A standard deviation is reported to three significant figures, so s = 0.400 mg/min; the full value stays in the calculator.

A full-credit answer: The standard deviation of the pH 8 readings is 0.400 mg/min.

Do not award the point for 0.16 (the quotient before the square root), 0.358 (dividing by 5 instead of n − 1) or 0.64 (the sum of the squared differences).

(c) Calculate the standard error of the pH 8 mean. (1 point)

mg/min

Write down the values in the question:

s = 0.400 mg/min

n = 5

Write down the equation:

S E_{𝑥̄} = \frac{s}{\sqrt{n}}

Substitute the values into the equation:

S E_{𝑥̄} = \frac{s}{\sqrt{n}}

S E_{𝑥̄} = \frac{0.400}{\sqrt{5}}

S E_{𝑥̄} = \frac{0.400}{2.236}

S E_{𝑥̄} = 0.179 mg/min

Round the value you report:

A standard error is reported to three significant figures, so SE = 0.179 mg/min.

A full-credit answer: The standard error of the pH 8 mean is 0.179 mg/min: the standard deviation, 0.400 mg/min, divided by the square root of the five tubes.

Do not award the point for 0.400 divided by 5 = 0.080 (dividing by n instead of the square root of n), for 0.400 divided by 2 = 0.200 (using the square root of n − 1), or for 0.400 itself.

(d) One student claims that the lipase works faster at pH 8 than at pH 6. Use ±2SE error bars for the two means to evaluate the claim, and state whether the null hypothesis of no difference between pH 6 and pH 8 is rejected. (1 point)

A full-credit answer: The pH 8 bar runs from 3.04 to 3.76 mg/min and the pH 6 bar from 2.36 to 2.84 mg/min. The top of the pH 6 bar, 2.84, sits below the bottom of the pH 8 bar, 3.04, so the two ±2SE bars do not overlap. The difference between the means is very unlikely to be chance: the data support the claim, and the null hypothesis of no difference is rejected.

𝑥̄ = 3.40 mg/min

SE = 0.179 mg/min

bar ends = 𝑥̄ \pm 2 S E

bar ends = 𝑥̄ \pm 2 S E

2 S E = 2 \times 0.179 = 0.358 mg/min

lower end = 3.40 - 0.358 = 3.04 mg/min

upper end = 3.40 + 0.358 = 3.76 mg/min

Check the box for each point your answer earns

the pH 8 bar runs from 3.40 − 0.358 = 3.04 to 3.40 + 0.358 = 3.76 mg/min (accept each end within ±0.01) and the pH 6 bar from 2.36 to 2.84; the bars do not overlap, so the difference is very unlikely to be chance, the claim is supported, and the null hypothesis is rejected.

Accept the pH 8 bar written as 3.40 ± 0.36 mg/min. Do not award the point for a decision made from the two means alone, or for a bar of ±1SE.

Common slip: Comparing the means alone, 3.40 against 2.60. The claim rests on the two ±2SE bars, which do not overlap; two means alone could differ by chance.

Free-response score: 0 of 4

Multiple choice checked: 0 of 18 correct.

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