S3 Statistics 3

Example:X₁ and X₂ are independent normal random variables

X₁   ~   N(21, 12)    and    X₂   ~   N(9, 6).

Find the expected mean and standard deviation of X₁ – 2X₂.

Solution:E[X₁] =21,  Var[X₁] = 12    and    E[X₂] = 9,  Var[X₂] = 6

Example:The weights of empty coffee jars are normally distributed with mean 0.1 kg and standard deviation 0.02 kg. The weight of coffee in the jars is normally distributed with mean 1 kg and standard deviation 0.06kg.

Find the distribution of 12 full jars of coffee.

What is the probability that 12 full jars weigh more than 13⋅5 kg?

Solution:Let X₁, X₂, ... X₁₂ be the weights of 12 empty jars and Y₁, Y₂, ... Y₁₂ be the weights of coffee in the jars.  X   ~   N(0⋅1, 0⋅02²)  and  Y   ~   N(1, 0⋅06²).

Let W be the total weight of 12 full jars then  W = X₁ + X₂ + ... + X₁₂ + Y₁ + Y₂ + ... + Y₁₂.

Then  E[W]   =   12 E[X] + 12 E[Y]   =   12 × 0⋅1 + 12 × 1   =   13⋅2

and, assuming independence,

Var[W]   =   12 Var[X] + 12 Var[Y]   =   12 × 0⋅02² + 12 × 0⋅06²   =   0⋅048.

As we are combining normal distributions

the distribution for 12 full jars is N(13.2, 0⋅048).  Answer

The probability that 12 full jars weigh more than 13⋅5 kg is

Example:How would you take a stratified sample of 50 children from a school of 500 pupils divided as follows:

Solution:As 50 is ¹/₁₀ of the total population,  ¹/₁₀ of each stratum should be selected in the sample.

Thus the sample would comprise

and simple random number sampling would be used within each stratum.

Example:A bag contains a large number of coins, of which 25 are 2p coins and 35 are 5p coins.

(a)X is the value of a single coin draw from the bag. Find the expected mean of all coins in the bag, µ = E[X].

Samples of size 3 are now drawn from the bag.

(b)Find the sampling distribution of X̅ and the expected value of (i) the median, and (ii) the mean.

(c)(i)The median, Q₂, is used as an estimator of the mean of all the coins, µ. Show that Q₂ is a biased estimator of µ, and find the bias.

(ii)The mean, X̅, is used as an estimator of the mean of all the coins, µ. Show that X̅ is an unbiased estimator of µ.

(d)kQ₂ is now used as an unbiased estimator of the mean of all the coins. Find the value of k.

Solution:

(a)µ = E[X] = ∑x_ip_i = 25 × 2 + 35 × 5 = 3∙8

(b)

Sampling distribution

(c)(i)The median Q₂ is used as the estimate of the mean.

From part (a) we know that the true value of the mean µ is 3⋅8, and in part (b) we have shown that E[Q₂] = 3⋅944

and the bias is |E[Q₂] – true value of µ| = |3⋅944 − 3⋅8| = 0⋅144

(ii) The mean X̅ is used as the estimate of the mean.

From part (a) we know that the true value of the mean µ is 3⋅8, and in part (b) we have shown that E[X̅] = 3⋅8

(d)If we now use kQ₂ as an unbiased estimator of the mean value of all the coins.

If kQ₂ is an unbiased estimator of µ, E[kQ₂] = true value of µ

Example:A sample of size 3 is drawn from a binomial distribution B(10, 0⋅25) and the mean, X̅, is calculated.

The probability of success, p, is estimated by p̂ = 110X. Show that p̂ is an unbiased estimator of p.

Solution:

For a sample {X₁, X₂, X₃},   X̅ = 13(X₁ + X₂ + X₃)

Example:A sample of size 4 is drawn from a continuous uniform distribution, U[3, β ]. The mean of the sample, X̅, is calculated.

The upper limit, β, is estimated by β̂ = 2X̅− 3. Show that β̂ is an unbiased estimator of β.

Solution:

For a sample {X₁, X₂, X₃, X₄},   X̅ = 14(X₁ + X₂ + X₃ + X₄)

which is the true value of the (unknown) upper limit, β.

Example:

The weights of a sample of five chocolate bars produced by a machine were 56, 53, 57, 51 and 54 grams. Find unbiased estimators for the weight of all chocolate bars produced by that machine.

Solution:

AnswerUnbiased estimators for the mean and variance of all chocolate bars are 54⋅2 grams and 5⋅7 grams².

Example: The volume of water in each of a sample of 14 litre bottles of water from a day’s production is taken. The results are shown below, in ml.

1023, 1019, 1004, 1011, 1023, 1014, 1017, 1020, 1020, 1010, 1025, 1007, 1016, 1019

Find unbiased estimates for the mean and variance of all bottles produced on that day.

Solution:

First find the sample mean,   X̅, = 1422814 = 1016⋅286….

(finding X − X̅ each time) would give unpleasant arithmetic,

so use (sd)_x² = 1n∑X² − X̅²

∑X² = 14460232

AnswerUnbiased estimators for the mean and variance of the whole day’s production are 1016⋅3 ml and 39⋅91 ml².

Example: The weights of a sample of 15 packets of biscuits are recorded and give the following results.

Σ X = 3797 grams, and Σ X ² = 973692.

Find unbiased estimators for the mean and variance of all biscuits produced by this process.

Solution:   µ = X̅= 379715 = 253 ∙133 … = 252 ∙1 grams.

AnswerUnbiased estimators are μ̂ =252⋅1 g, and σ̂² = 896⋅1 g².

Example: The lengths of 10 rods are measured, and the sample has mean, X̅ = 26⋅7 cm and variance s² = 76⋅9 cm². An eleventh rod has length 30 cm.

Find (a) the mean and (b) the variance of the sample of 11 rods.

Solution: (a)   With the sample mean there are no complications.

For n = 10,   X̅₁₀ = 11010∑i=1X_i = 26 ∙7 ⇒ 10∑i=1X_i = 267

For n = 11,   11∑i=1X_i = 267 + 30 = 297 ⇒ X₁₁ = 11111∑i=1X_i = 29711 = 27 cm

(b)WARNING: The question refers to the variance of the sample, which means the unbiased estimate of the variance of the population.

For 11 rods, sample mean is 27 cm, and sample variance is 70⋅2 cm².

Example:A sample of size 50 is taken from a population of eggs with mean 23⋅4 grams and variance 36 grams².

(i)Find the probability that a single egg weighs more than 25 grams.

(ii)Find the probability that the sample mean is larger than 25.

(iii)What assumptions did you make?

Solution:

(i)The weight of a single egg, X ∼ N(23⋅4, 6²)

(ii)µ = 23⋅4,   σ² = 36

The sample mean X̅ ∼ N(23⋅4, ( 6√50 )²)

= 1 – Φ(1⋅89) = 1 – 0⋅9706   (from Normal tables)

= 0 ⋅0294.

(iii)We have assumed the Central Limit Theorem: in particular that the sample means form a normal distribution.

Example:A biscuit manufacturer makes packets of biscuits with a nominal weight of 250 grams. It is known that over a long period the variance of the weights of the packets of biscuits produced is 25 grams². A sample of 10 packets is taken and found to have a mean weight of 253⋅4 grams. Find 95% confidence limits for the mean weight of all packets produced by the machine.

Solution:

First assume that the machine is still producing packets with the same variance, 25.

Suppose that the mean weight of all packets of biscuits is µ grams then the population of all packets has mean µ and standard deviation 5.

From the central limit theorem we can assume that the sample means form an approximately normal population with mean µ and standard error (standard deviation) σ√n = 5√10 = 1⋅5811

95% of the samples will have a mean in the region

We assume that the mean of this sample, 253⋅4, lies in this region

This means that 95% of the samples will give an interval which contains the mean

and we say that  [250⋅3 g, 256⋅5 g]  is a  95% confidence interval  for µ.

This means that there is a 0⋅95 probability that this interval contains the true mean.

It does not mean that there is a probability of 0⋅95 that the true mean lies in this interval - the true mean is a fixed number, and either does or does not lie in the interval so the probability that the true mean lies in the interval is either 1 or 0.

Example:A sample of 64 packets of cornflakes has a mean weight X̅ = 510 grams and a variance S² = 36 grams². Find 90% confidence limits for the mean weight of all packets.

(Note that the ‘sample variance’ is taken as the unbiased estimate of µ².)

Solution:

We assume that the sample variance = the variance of the population of all packets

Now find standard deviation (standard error) of the sampling distribution of the mean (population of sample means),

For 90% confidence limits  z = ± 1⋅6449  (remember to use the 4 D.P. tables after the Normal Dist. tables),

using the sample mean X̅ = 510 grams

Note that we have assumed that the unbiased estimate, S² (=36), is the actual variance, µ², of the population.

This is a reasonable assumption as the number in the sample, 64, is large and the error introduced is therefore small.

Example:A machine, when correctly set, is known to produce ball bearings with a mean weight of 84 grams with a standard deviation of 5 grams. The production manager decides to test whether the machine is working correctly and takes a sample of 120 ball bearings. The sample has mean weight 83.2 grams. Would you advise the production manager to alter the setting of his machine? Use a 5% significance level.

Solution:

1)H₀:  µ = 84 grams

2)H₁:  µ ≠ 84 grams⇒ 2 tail test

(Note that the machine is not working correctly if the test result is too high or too low)

3)5% Significance level

4)The Test

We assume that the machine is still working with a standard deviation of σ = 5 g.

From H₀, the mean weight of all ball bearings is assumed to be µ = 84 g.

These are the parameters for the population of all ball bearings.

We want to test a sample mean and therefore need the mean and standard deviation of the population of sample means  (the sampling distribution of the sample mean, X̅).

and

We have an observed mean of 83⋅2

For a two-tailed test at 5%, we take 2⋅5% at

each end

   =  (1 – Φ(1⋅75)) = 0⋅0401

   = 4⋅01%   >   2⋅5%

and so not significant at the 5% level.

5)Conclusion

Do not reject H₀ at the 5% level and advise the production manager that there is evidence that he should not change his setting, or that there is evidence that the machine is working correctly, etc.

Example:The weights of chocolate bars produced by two machines, A and B, are known to be normally distributed with variances σ_A² = 4 and σ_B² = 3 grams². Samples are taken from each machine of sizes n_A = 25 and n_B = 16 which have means X̅_A = 123⋅1 and X̅_B = 124⋅4 grams. Is there any evidence at the 5% significance level that the bars produced by machine B are heavier than the bars produced by machine A?

Solution:

Suppose that the mean weights for all bars from the two machines are µ_A and µ_B

H₀:µ_A = µ_B

H₁:µ_B > µ_A  one-tail test  at 5% level

The test statistic is the observed difference between sample means,  X̅_B − X̅_A = 124⋅4 – 123⋅1 = 1⋅3,

and we must find the variance of this population of differences of sample means (the sampling distribution of differences of sample means).

Consider the population of differences of sample means  X̅_B − X̅_A.

Firstly, for the population of sample means for machine B

and secondly, for the population of sample means for machine A

and so for the population of differences of sample means

The observed difference, the test statistic, is 124⋅4 – 123⋅1 = 1⋅3

The Central Limit Theorem tells us that we have a Normal distribution

= 1 – Φ(2⋅2053)

=  1 – Φ(2⋅20)  =  1 – 0⋅9861  =  0⋅0134

= 1⋅34%  <  5%

Fortunately  (!)  the formula for testing the difference between sample means

is in your formula booklet.

Example:A machine usually produces steel rods with a mean length of 25⋅4 cm. The production manager wants to test 80 rods to see whether the machine is working correctly. The sample has mean 25⋅31 cm and variance 0⋅332 cm². Advise the production manager, using a 5% level of significance.

Solution:

H₀:µ = 25⋅4.

H₁:µ≠ 25⋅4  two-tail test, 2⋅5% in each tail

We assume that population variance  σ²  =  the sample variance  S² = 0⋅332

For the population of sample means (the sampling distribution of the sample means)

and

The observed sample mean is 25.31 and for a two-tail test at 5% we consider

Example:A firm has two machines, A and B, which make steel cable. 40 cables produced by machine A have a mean breaking strain of  1728 N and variance of 75² N², whereas 65 cables produced by machine B have a mean breaking strain of 1757 N and a variance of 63² N². Is there any evidence, at the 10% level, to suggest that machine B is producing stronger cables than machine A?

Solution:

Let  µ_A and  µ_B be the mean breaking strengths of all cables produced by machines  A  and B.

1)H₀:µ_A = µ_B

2)H₁:µ_B > µ_A  1 tail test

3)Significance Level 10%.

4)The Test

For Machine A

We assume that the population variance, σ_A² = the sample variance, S_A² = 75²

For Machine B

We assume that the population variance, σ_B² = the sample variance, S_B² = 63²

For differences in sample means  X̅_B − X̅_A

We have an observed difference in means,

test statistic,  X̅_B − X̅_A = 1757 – 1728 = 29

and for a 1-tail test that B is stronger

we need the area to the right of  29     mean is treated as continuous, so do not use 28.5

which is significant at 10%.

5)Conclusion

Reject H₀ at the  10% level and conclude that there is evidence that machine B produces cables with a greater mean strength than machine A.

Example:A die is rolled 300 times and the frequency of each score recorded.

Test whether the die is fair at the 2⋅5% level of significance.

Solution:

H₀:The die is fair, the probability of each score is ¹/₆.

H₁:The die is not fair, the probability of each score is not ¹/₆.

The expected frequencies are all  ¹/₆× 300 = 50 and we have

and  ν = number of degrees of freedom  =  n – 1  = 6 – 1 = 5

since the total is a linear equation connecting the frequencies and is fixed.

From tables we see that    χ²₅(2⋅5%) = 12⋅832 > 2⋅64 , so our observed result is not significant.

We do not reject H₀  and conclude that the die is fair.

Example:A switchboard operator records the number of new calls in 69 consecutive one-minute periods in the table below.

a)Say why you think that a Poisson distribution might be suitable.

b)Find the mean and variance of this distribution. Do these figures support the view that they might form a Poisson distribution?

c)Test the goodness of fit of a Poisson distribution at the 5% level.

Solution:

a)Telephone calls are likely to occur singly, randomly, independently and uniformly which are the conditions for a Poisson distribution.

b)Treating  ≥ 6   as 7 we calculate the mean and variance

From these figures we can see that the mean and variance are approximately equal: since the mean and variance of a Poisson distribution are equal this confirms the view that the distribution could be Poisson.

c)H₀:The Poisson distribution is a suitable model

H₁: The Poisson distribution is not a suitable model.

The Poisson probabilities can be calculated from   P(r) = λ^re^−λr!    where λ = 3⋅12, and the expected frequencies by multiplying by N = 69.

Note that the probability for  ≥ 6 is found by adding the other probabilities and subtracting from 1.

The expected frequency for x = 0   is  3.06 < 5   so it has been grouped with x = 1.

Thus we have n = 6 classes (after grouping) and   ν  = n – 2 = 4

We have calculated  χ² =  1.92  <  9.488  which is not significant so we do not reject H₀

and conclude that the Poisson distribution is a suitable model.

Example:The sizes of men’s shoes purchased from a shoe shop in one week are recorded below.

Is the manager’s assumption that the normal distribution is a suitable model justified at the 5% level?

Solution:

H₀:The normal distribution is a suitable model

H₁:The normal distribution is not a suitable model.

The total number of pairs, mean and standard deviation are calculated to be   175,  8.886  and  1.713  (taking ≤ 6  as 5   and   ≥ 12  as 12)

Remembering that   size 8   means from 7.5 to 8.5 we need to find the area between 7.5 and 8.5 and multiply by 175 to find the expected frequency for size 8, and similarly for other sizes.

n = 7 classes & 3 linear equations connecting the frequencies (N, m ,s)  ⇒ν = n – 3 = 4.

χ²₄(5%) = 9.488   and we have calculated  χ² = 5.34 < 9.488 and so we do not reject H₀   and therefore conclude that the normal distribution is a suitable model.

Example:Natives of England, Africa and China were classified according to blood group giving the following table.

Is there any evidence at the 5% level that there is a connection between blood group and nationality?

Solution:

H₀:There is no connection between blood group and nationality.

H₁:There is a connection between blood group and nationality.

First redraw the table showing totals of each row and column

Now we need to calculate the expected frequency for English and group O. There are 609 English and 1335 people altogether so  ⁶⁰⁹/₁₃₃₅   of the people are English, and from H₀ we know that there is no connection between blood group and nationality, so there should be  ⁶⁰⁹/₁₃₃₅   of those with group O who are also English

this can become automatic if you notice that you just multiply the totals for the row and column concerned and divide by the total number

The value of χ² is calculated below

We have   ν = (4 – 1)(3 – 1) = 6   degrees of freedom   and   χ²₆(5%) = 12.592 .

We have calculated  χ² = 8.41  < 12.592

⇒   do not reject   H₀   and therefore conclude that there is no connection between   nationality and blood group.

Example:Rank  the following numbers:  45, 65, 76, 56, 34, 45, 23, 67, 65, 45, 81, 32.

Solution:First put in order and give ranks as if all were different: then give the average rank for those which are equal.

You must now calculate the PMCC, not Spearman, using the modified ranks.

Example:Ten varieties of coffee labelled  A, B, C, ..., J  were tasted by a man and a woman. Each ranked the coffees from best to worst as shown.

Find Spearman’s rank correlation coefficient.

Solution:Rank for each person, find d and then r_s.

Example:The product moment correlation coefficient between 40 pairs of values is +0.52. Is there any evidence of correlation between the pairs at the 5% level?

Solution:

H₀:There is no correlation between the pairs,  ρ = 0.

H₁:There is correlation, positive or negative, between the pairs,   ρ ≠ 0, two-tail test

From tables for n = 40 which give one-tail figures, we must look at the 2.5% column and the critical values are  ±0.3120

The calculated figure is  0.52 > 0.3120  and so is significant

⇒ we reject  H₀  and conclude that there is some correlation (positive or negative) between the pairs.

Example:It is believed that a person who absorbs a drug well on one occasion will also absorb a drug well on another occasion. Tests on ten patients to find the percentage of drug absorbed gave the following value for Spearman’s rank correlation coefficient,  r_s = 0.634. Is there any evidence at the 5% level of a positive correlation between the two sets of results.

Solution:

H₀:There is no correlation between the two sets of results,  ρ_s = 0,

H₁:There is positive correlation between the two sets of results,  ρ_s > 0, one-tail test.

From the tables for n = 10 and a one-tail test the critical value for 5% is  0.5364.

The calculated value is 0.634  >  0.5364  which is significant

⇒ reject H₀; conclude that there is evidence of positive correlation between the two sets of results.

Note that this shows correlation between the ranks of the two sets of results.

Example:A random sample of 8 students sat examinations in Geography and Statistics. The product moment correlation coefficient between their results was 0⋅572 and the Spearman rank correlation coefficient was 0⋅655.

(a)  Test both of these values for positive correlation. Use a 5% level of significance.

(b)  Comment on your results.

Solution:

(a)H₀ :  ρ = 0  ;  H₁ :  ρ > 0

For the PMCC

the 5% Critical Value is  0⋅6215

0⋅572 < 0⋅6215      ⇒        not significant at 5%

⇒     there is evidence that there is no positive correlation.

For Spearman’s rank correlation coefficient

the 5% Critical Value is  0⋅6429

0⋅655 > 0⋅6429      ⇒        significant at 5%

⇒     there is evidence of positive correlation.

(b)From the PMCC there is not enough evidence to conclude that as Statistics marks increased Geography marks also increased

− i.e. conclude that the points on a scatter diagram do not lie close to a straight line.

From Spearman’s rank correlation coefficient there is evidence that students ranked highly in Statistics were also ranked highly in Geography, or people with high scores in Statistics also had high scores in Geography

Example:A bag has 468 beads of two colours, white and green. 20 beads are taken at random and the number, i, of green beads in the sample is counted.

To estimate the true number of green beads, g, in the bag, we calculate

If g is the true number of green beads in the bag

thenthe probability of drawing a green bead in a single trial is  p = g468 ,

anddrawing  n = 20  beads with replacement gives a Binomial distribution  B (n, p).

Thusµ = E[i] = np = 20 × g468

We do not actually know the number of green beads, and want to estimate this number after taking one sample

estimateĝ = i20 × 468.

We now find the expected value of this estimate

Example

A naturalist wishes to estimate the number of squirrels in a wood. He first catches 50 squirrels, marks them and then releases them. Later he catches 30 squirrels and counts the number, i, which have been marked.

The true number in the population, n, is then estimated as n̂ from the equation

i.e.it is possible that i = 0, in which case the estimate  n̂  is infinite   when  i = 0,

Example:A large bag contains counters: 60% have the number 0, and 40% have the number 1.

(a)Find the mean, µ, and variance, σ².

A simple random sample of size 3 is drawn.

(b)List all possible samples.

(c)Find the sampling distribution for the mean  X̅ = X₁ + X₂ + X₃3

(d)Use your answers to part (c) to find  E[X̅], and  Var[X̅].

(e)Find the sampling distribution for the mode  M.

(f)Use your answers to part (e) to find  E[M], and  Var[M].

Solution:

(a)µ = ∑ x_ip_i = 0 × 0⋅6  +  1 × 0⋅4  =  0⋅4

σ² = ∑ x_i²p_i – µ² = (0² × 0⋅6  +  1² × 0⋅4)  –  0⋅4²  =  0⋅24

(b)Possible samples are

(0, 0, 0)     (1, 0, 0)     (1, 1, 0)     (1, 1, 1)

           (0, 1, 0)     (1, 0, 1)

           (0, 0, 1)     (0, 1, 1)

(c)From (c) we can find the sampling distribution of the mean

(d)E[X̅] = 0 × 0⋅216  +  13 × 0⋅432  +  23 × 0⋅288  +  1 × 0⋅064  =  0⋅4

Var[X̅] = (0² × 0⋅216  +  13² × 0⋅432  +  23² × 0⋅288  +  1² × 0⋅064)  – 0⋅4²

(e)From (c) we can find the sampling distribution of the mode

(f)E[M] = 0 × 0⋅648  +  1 × 0⋅352  =  0⋅352

Var[M] = (0² × 0⋅648 + 1² × 0⋅352) – 0⋅352²