1. “Teach A Level Maths” Yr1/AS Statistics Sample 1
© Christine Crisp

2. Part 2 of “Teach A level Maths” covers the work on Statistics for Year 1/AS of the Maths A level.
All topics for the 4 specifications offered by the English examining bodies are covered. Where a topic relates to some specifications only, this is indicated in a contents file and also at the start of the presentation.

3. Explanation of Clip-art images
An important result, example or summary that students might want to note.
It would be a good idea for students to check they can use their calculators correctly to get the result shown.
An exercise for students to do without help.

4. The slides that follow are extracts from 6 of the 24 presentations on Statistics.
Random Samples
Cumulative Frequency Diagrams
Scatter and Correlation
Histograms
Binomial Problems
Hypothesis Testing

5. Random Samples
This extract is from a very early presentation and is part of a discussion of stratified sampling.

6. However, even a random sample can be biased.
e.g. Consider again our earlier example of wanting to know how many of 1200 students are vegetarian if we know there are equal numbers of boys and girls.
Decide with your partner how you could avoid gender bias in taking a sample of 40.
Ans: Take 2 random samples, one of 20 boys and one of 20 girls, and then combine them.
This is a stratified sample.
The word stratified comes from strata * which means layers. The sample is formed from 2 layers, boys and girls.
* One stratum but two or more strata.

7. However, we are unlikely to have exactly equal numbers of boys and girls.
Can you see what to do if the college has 660 boys and 540 girls and we need a sample of 40 ?
Ans: We sample in proportion to the numbers in the strata.
660
Boys:  40
= 22
1200
Girls:
We find the number in the final stratum by subtracting from the total sample size:
40 – 22 = 18

8. 660 50 710 540 70 610 1200 120 1320 We may have more than 2 strata.
Suppose we include staff in the survey.
Students
Staff
Total
Male
660 50
710
Female
540 70
610
1200
120
1320
The sample is to be stratified for both gender and student/staff.
How many male staff should be in a sample of 40 ?
Solution:
There are 50 male staff, so we need

9. 660 50 710 540 70 610 1200 120 1320 50 We may have more than 2 strata.
Suppose we include staff in the survey.
Students
Staff
Total
Male
660 50
710
Female
540 70
610
1200
120
1320
The sample is to be stratified for both gender and student/staff.
How many male staff should be in a sample of 40 ?
Solution:
There are 50 male staff, so we need 50

10. We may have more than 2 strata.
Suppose we include staff in the survey.
Students
Staff
Total
Male
660 50
710
Female
540 70
610
1200
120
1320
The sample is to be stratified for both gender and student/staff.
How many male staff should be in a sample of 40 ?
Solution:
There are 50 male staff, so we need 50  40 =
1·51…
1320
We must have a whole number so we round to 2.

11. Cumulative Frequency Diagrams
Histograms
Cumulative Frequency Diagrams
As well as explaining theory, the presentations show worked examples and set introductory exercises.
This exercise is a simple example to introduce students to using Excel to draw a statistical diagram. Solutions are given after the exercises.

12. N.B. The link cannot be used in this sample presentation.
Exercise
The table and diagram show the number of flowers in a sample of 43 antirrhinum plants. x
U.C.B. f
Cu. f
0 - 19 6
40 – 59 10
60 – 79 12
80 – 99 7 5 1
Either: Draw the diagram on squared paper.
Or: Use EXCEL with a Scatter diagram. Select the box below to link to Excel.
N.B. The link cannot be used in this sample presentation.
Estimate from the diagram the median number of plants and the percentage of plants that have more than 90 flowers.
Source: Statistics for Biology by O N Bishop published by Pearson Education
Antirrhinums

13. There are 43 observations, so the median is given by the 21·5th one.
Solution
U.C.B. f
Cu. f
19.5
39.5 6
59.5 10 16
79.5 12 28
99.5 7 35
119.5 5 40
139.5 1 41
159.5 42
179.5 43 32
There are 43 observations, so the median is given by the 21·5th one.
Median = 69 (approx.)
Estimated number with more than 90 flowers =

14. Solution
U.C.B. f
Cu. f
19.5
39.5 6
59.5 10 16
79.5 12 28
99.5 7 35
119.5 5 40
139.5 1 41
159.5 42
179.5 43 32
There are 43 observations, so the median is given by the 21·5th one.
Median = 69 (approx.)
Estimated number with more than 90 flowers =
43 – 32 = 11
26%
Percentage with more than 90 flowers

15. Scatter and Correlations
Students need to understand Correlation and be able to draw and interpret scatter diagrams. Causal correlation is covered in the same presentation.
This extract links again to Excel with help for drawing a diagram with simple data.

16. Children's Accidents (%), y
Exercise
2. The table shows the percentage of open spaces in 9 areas of London, x, together with the number of accidents to children as a percentage of those to adults, y. A B C D E F G H I
Open Spaces (%), x
5.0
1.3
1.4
7.0
4.5
5.2
6.3
14.6
14.8
Children's Accidents (%), y
46.3
42.9
40.0
38.2
37.0
33.6
30.8
23.8
17.1
If you know how to use a spreadsheet to draw diagrams, copy and paste special the sample into a spreadsheet (using HTML) and choose a scatter diagram to display it.
If you need help to do the exercise, select the option here.
The link cannot be used in this sample. There is a live link in Sample 2.
London

17. The next slide shows a screen shot of the Excel sheet with the instructions and the diagram the students will produce before returning to the Powerpoint presentation for the question and solution.

19. Children's Accidents (%), y
Exercise
2. The table shows the percentage of open spaces in 9 areas of London, x, together with the number of accidents to children as a percentage of those to adults, y. A B C D E F G H I
Open Spaces (%), x
5.0
1.3
1.4
7.0
4.5
5.2
6.3
14.6
14.8
Children's Accidents (%), y
46.3
42.9
40.0
38.2
37.0
33.6
30.8
23.8
17.1
Describe the correlation and estimate the number of accidents if 10% of the area is open space.
The diagram and solution are on the next slide.

20. Solution: Exercise 28%. There is a strong, negative correlation.
There is evidence that the percentage of children’s accidents decreases as the percentage of open space in areas of London increases.
If 10% of the area is open space, we estimate the number of accidents to be. . .
28%.

21. Histograms Histograms
Students need to understand the difference between a Bar Chart and a Histogram.
The exercise shown here reinforces the rules without the students needing to spend time drawing a diagram.

22. Histograms
Exercise
95 components are tested until they fail. The table gives the times taken ( hours ) until failure.
Time to failure (hours)
0-19
20-29
30-39
40-44
45-49
50-59
60-89
Number of components 5 8 16 22 18 10
Find 3 things wrong with the histogram which represents the data in the table.

23. Histograms Answer: 0-19 20-29 30-39 40-44 45-49 50-59 60-89 5 8 16 22
Time to failure (hours)
0-19
20-29
30-39
40-44
45-49
50-59
60-89
Number of components 5 8 16 22 18 10
Frequency has been plotted instead of frequency density.
There is no title.
There are no units on the x-axis.

24. Histograms Incorrect diagram Time taken for 95 components to fail

25. Binomial Problems
A vital part of the study of the Binomial model is for students to understand when it is appropriate to use it.
The following extract shows a worked example with a discussion of the conditions needed for the model to be used.

26. e. g. 1. A factory produces a particular type of computer chip
e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect?
Are the conditions met for using the Binomial model?
A trial has 2 possible outcomes, success and failure.
Yes: Each chip is either defective or not.
The trial is repeated n times.
Yes: 20 chips are selected so n = 20.
The probability of success in one trial is p and p is constant for all the trials.
Yes: We are given 15% ( from which we can find p ) and we can assume it is constant.
The trials are independent.
Yes: The probability of selecting a defective chip does not depend on whether one has already been selected.

27. e. g. 1. A factory produces a particular type of computer chip
e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect?
Solution:
Let X be the r.v. “number of defective chips”
We must never miss out this stage since it reminds us that
(i) X represents a number ( that can be 0, 1, 2, n ), and
(ii) we have to make the decision as to whether to count the number of defective chips or perfect ones.
So,
Writing the distribution of X in this way makes us check that we have the p that fits our definition of the r.v.: defective rather than perfect.

28. e. g. 1. A factory produces a particular type of computer chip
e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect?
Solution:
Let X be the r.v. “number of defective chips”
So,
The solution is now straightforward. We want
We need to be very careful here and not use by mistake.
I had set up the Binomial for the number of defective chips, because I had the proportion for defective. However, the question asked for the probability of 19 perfect ones.
If I had written
Let X be the r.v. “ number of perfect chips”
Then,
and I would want

29. Hypothesis Testing
The presentation covers both one-tailed and two-tailed tests.
The following slides are the first few introducing the topic. The worked example is completed and students are shown how to write out a clear solution before continuing to an exercise and further theory.

30. Suppose there is a new drug treatment which we hope will be better than the existing one.
To test whether it is better, we could set up a trial involving a certain number of people. We could then see how long it takes people to get well with the new drug and compare it with results for the old one.
However, sampling involves random effects and we need to know whether the apparently good results are really due to improvements in treatment.
Statistics has a big part to play in making decisions of this type and this presentation introduces an important theory that is widely used.

31. 75% of patients improved compared with 60% before.
e.g. 1. In a trial of 20 patients with a new drug, the condition of 15 was greatly improved. On the older drug, 60% reported the same improvement. Is there evidence that the new drug is more effective than the old one?
We can’t just say that the drug is clearly better because
75% of patients improved compared with 60% before.
We need to find out what the probability is that 15 patients improve even if the new drug is no more effective than the old one.
The situation can be modelled with the Binomial distribution.
We let p, the probability of success, be the probability that a patient improved with the new drug.

32. H0 is called the null hypothesis. ( I think of it as “no change”. )
We set up a hypothesis ( a theory ) that the new drug is not more effective and only reject this if we have significant evidence against it.
We write
H0 is called the null hypothesis. ( I think of it as “no change”. )
The null hypothesis is using the value of p from data for the old drug.
We want to test if the new drug is better than the old one so we also have
H1 is the alternative hypothesis.

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