Stats for Engineers Lecture 7

Confidence intervals

During component manufacture, a random sample of 500 are weighed each day, and each day a 95% confidence interval is calculated for the mean weight of the components. On the first day we obtain a confidence interval for the mean weight of Kg. Which of the following is correct on average if the (unknown) mean weight and (known) standard deviation remain constant on different days?

1 2 3 4

74%

13%13%

0%

1. On 95% of days, the mean weight is in the range Kg

2. 95% of the daily sample means lie in the range Kg

3. On 95% of days the calculated confidence interval contains

4. 95% of the components have weights in the range Kg

A 95% confidence interval for if we measure a sample mean and already know is

Recap: Confidence Intervals for the mean

Random sample from , where is known [or large data sample so can estimate it accurately, ] but is unknown.

We want a confidence interval for .

Reminder:

With probability 0.95, a Normal random variables lies within 1.96 standard deviations of the mean.

95% of the time expect in

P=0.025P=0.025

𝑋=𝜇 ±1.96 √ 𝜎2

𝑛 ⇒𝜇=𝑋 ±√ 𝜎2

𝑛

Confidence interval interpretation

During component manufacture, a random sample of 500 are weighed each day, and each day a 95% confidence interval is calculated for the mean weight of the components. On the first day we obtain a confidence interval for the mean weight of Kg. Which of the following are correct on average if the (unknown) mean weight and (known) standard deviation remain constant on different days?

, so 95% of the time the sample mean lies in

Every day we get a different sample, so different .

Day 1: Day 2: Day 3: Day 4: …

e.g. Confidence interval

…

95% of the time, is in 95% of the time, is in

here

On 95% of days, the mean weight is in the range Kg

95% of the daily sample means lie in the range Kg

95% of the components have weights in the range Kg

Other answers?

NO – the confidence interval we calculated is for the mean weight, not individual weights(and so the mid-point is incorrect)

NO – the correct statement is that 95% of the time lies in Statement would only be true if on the first day we got , which has negligible probability

NO - the mean weight is assumed to be a constant. It is either in the range or it isn’t – if true for one day it will be true for all days

A confidence interval for if we measure a sample mean and already know is where

In general can use any confidence level, not just 95%.

95% confidence level has 5% in the tails, i.e. p=0.05 in the tails.

In general to have probability in the tails; for two tail, in each tail:

p/2p/2

Q

E.g. for a 99% confidence interval, we would want .

Two tail versus one tail Does the distribution have two small tails or one? Or are we only interested in upper or lower limits?

If the distribution is one sided, or we want upper or lower limits, a one tail interval may be more appropriate.

P=0.05

95% One tail

P=0.025P=0.025

95% Two tail

Example

You are responsible for calculating average extra-urban fuel efficiency figures for new cars. You test a sample of 100 cars, and find a sample mean of The standard deviation is . What is the 95% confidence interval for the average fuel efficiency?

Answer:

Sample size if and 95% confidence interval is .

⇒55.4 ±1.96√ 1.22100=55.4 ±0.235

i.e. mean in 55.165 to 55.63 mpg at 95% confidence

1 2 3 4

0% 0%0%0%

Confidence interval

Given the confidence interval just constructed, it is correct to say that approximately 95% of new cars will have efficiencies between 55.165 and 55.63 mpg?

Question from Derek Bruff

1. YES – high confidence2. YES – low confidence3. NO – high confidence4. NO – low confidence

NO: mpg given in the question is the standard deviation of the individual car efficiencies (i.e. expect new cars in a range . The confidence interval we calculated is the range we expect the mean efficiency to lie in (much smaller range).

Countdown

10

Example: Polling

A sample of 1000 random voters were polled, with 350 saying they will vote for the Conservatives and 650 saying another party. What is the 95% confidence interval for the Conservative share of the vote?

Answer: this is Binomial data, but large so can approximate as Normal

𝜎 2=𝑛𝑝 (1−𝑝 )≈1000×0.35× (1−0.35 )=227.5

Random variable is the number voting Conservative,

Take variance from the Binomial result with

⇒𝜎=√227.5≈15.195% confidence interval for the total votes is

95% confidence interval for the fraction of the votes is

350±29.61000 ≈0.35±0.03 i.e. 3% confidence interval

Example – variance unknown

A large number of steel plates will be used to build a ship. A sample of ten are tested and found to have sample mean and sample variance What is the 95% confidence interval for the mean weight ?

Reminder:Sample Variance:

Normal data, variance unknown Random sample from , where a are both unknown.

Want a confidence interval for , using observed sample mean and variance.

When we know the variance: use which is normally distributed

But don’t know , so have to use sample estimate

When we don’t know the variance: use which has a t-distribution (with d.o.f)

Sometimes more fully as “Student’s t-distribution”

Wikipedia

Remember:

𝜈=𝑛−1=1 𝜈=𝑛−1=5 𝜈=𝑛−1=50

Normal

t-distribution

For large the t-distribution tends to the Normal - in general broader tails

Confidence Intervals for the mean

If is known, confidence interval for is to , where is obtained from Normal tables (z=1.96 for two-tailed 95% confidence limit).

If is unknown, we need to make two changes:

(i) Estimate by , the sample variance;

(ii) replace z by , the value obtained from t-tables,

The confidence interval for if we measure a sample mean and sample variance is: to .

t-tables

give for different values Q of the cumulative Student's t-distributions, and for different values of

Q

𝑄 (𝑡𝜈)=∫−∞

𝑡𝜈

𝑓 𝜈 (𝑡 )𝑑𝑡

The parameter is called the number of degrees of freedom.

(when the mean and variance are unknown, there are degrees of freedom to estimate the variance)

Q

For a 95% confidence interval, we want the middle 95% region, so Q = 0.975 (0.05/2=0.025 in both tails).

Similarly, for a 99% confidence interval, we would want Q = 0.995.

t-distribution example:

A large number of steel plates will be used to build a ship. A sample of ten are tested and found to have sample mean and sample variance What is the 95% confidence interval for the mean weight ?

From t-tables, for Q = 0.975

Answer:

95% confidence interval for is:

i.e. 1.95 to 2.31

= 2.2622.

⇒𝜇=2.13±2.2622√ 0.25210=(2.13 ±0.18 ) kg

1 2 3 4

5%

14%

33%

48%

Confidence interval width

We constructed a 95% confidence interval for the mean using a random sample of size n = 10 with sample mean . Which of the following conditions would NOT probably lead to a narrower confidence interval?

Question adapted from Derek Bruff

1. If you decreased your confidence level

2. If you increased your sample size 3. If the sample mean was smaller4. If the population standard

deviation was smaller

95% confidence interval for is:; width is

Confidence interval width

We constructed a 95% confidence interval for the mean using a random sample of size n = 10 with sample mean . Which of the following conditions would NOT probably lead to a narrower confidence interval?

Decrease your confidence level?

larger tail smaller smaller confidence interval

Increase your sample size?

Smaller sample mean?

Smaller population standard deviation?

larger smaller confidence interval ( and both likely to be smaller)

smaller just changes mid-point, not width

likely to be smaller smaller confidence interval

Sample size

How many random samples do you need to reach desired level of precision?

For example, for Normal data, confidence interval for is .

Suppose we want to estimate to within , where (and the degree of confidence) is given.

⇒𝑛=𝑡𝑛− 12 𝑠2

𝛿2

Want 𝛿=𝑡𝑛−1√ 𝑠2𝑛Need: - Estimate of (e.g. previous experiments)

- Estimate of . This depends on n, but not very strongly.

e.g. take for 95% confidence.

Rule of thumb: for 95% confidence, choose

Example

A large number of steel plates will be used to build a ship. Ten are tested and found to have sample mean weight and sample variance How many need to be tested to determine the mean weight with 95% confidence to within ?

Answer:

⇒𝑛=𝑡𝑛− 12 𝑠2

𝛿2=2.1

20.252

0.12=27.6

Want 𝛿=0.1 kg=𝑡𝑛− 1√ 𝑠2𝑛Take for 95% confidence.

i.e. need to test about 28

1. 2 3 4

17% 17%

28%

39%

Number of samples

If you need 28 samples for the confidence interval to be approximately how many samples would you need to get a more accurate answer with confidence interval

1. 88.52. 2803. 28004. 28000

𝛿=𝑡𝑛−1√ 𝑠2𝑛 so need more. i.e. 2800

Linear regression

Linear regression: fitting a straight line to the mean value of as a function of

403020100

250

200

150

100

x

y

We measure a response variable at various values of a controlled variable

e.g. measure fuel efficiency at various values of an experimentally controlled external temperature

𝑥

𝑦

𝑥1 𝑥2 𝑥3

Distribution of when

Regression curve: fits the mean values of the distributions

𝑦=𝑎𝑥+𝑏

403020100

250

200

150

100

x

y

From a sample of values at various , we want to fit the regression curve.

e.g.

403020100

250

200

150

100

x

y

Or is it

What do we mean by a line being a ‘good fit’?

Straight line plots

Which graph is of the line ?

1 2 3 4

56%

0%

33%

11%

1. Plot2. Plot23. Plot34. Plot4

1. 2.

3. 4.

𝑥

𝑦

Simple model for data:

𝑦 𝑖=𝑎+𝑏𝑥 𝑖+𝑒𝑖

Equation of straight line is

Random errorStraight line

- Linear regression model

Simplest assumption: for all , and 's are independent

Maximum likelihood estimate = least -squares estimate

Minimize 𝐸=∑𝑖𝑒𝑖2=¿∑

𝑖(𝑦 𝑖− �̂� 𝑖)=∑

𝑖( 𝑦 𝑖−𝑎−𝑏𝑥 𝑖 )

2¿

Data point Straight-line prediction

Model is

E is defined and can be minimized even when errors not Normal – least-squares is simple general prescription for fitting a straight line

(but statistical interpretation in general less clear)

Want to estimate parameters a and b, using the data.

e.g. - choose and to minimize the errors

The line has been proposed as a line of best fit for the following four sets of data. For which data set is this line the best fit (minimum )?

1 2 3 4

57%

0%

24%19%

Question from Derek Bruff

1. Pic12. Pic23. Pic correct4. pic4

1. 2.

3. 4.

How to find and that minimize ?

For minimum want and , see notes for derivation

Solution is the least-squares estimates and :

�̂�=𝑆𝑥𝑦

𝑆𝑥𝑥∧�̂�=𝑦− �̂�𝑥

𝑆𝑥𝑥=∑𝑖𝑥 𝑖2−

(∑𝑖 𝑥𝑖)2

𝑛 =∑𝑖

(𝑥𝑖−𝑥 )2

𝑆𝑥𝑦=∑𝑖𝑥 𝑖 𝑦 𝑖−

∑𝑖𝑥 𝑖∑

𝑖𝑦 𝑖

𝑛 =∑𝑖

(𝑥𝑖−𝑥) ( 𝑦 𝑖− 𝑦 )❑

Where

Sample means

�̂�=�̂�+�̂� 𝑥Equation of the fitted line is

Note that since

⇒ �̂�−𝑦=�̂� (𝑥−𝑥)

i.e. is on the line

403020100

250

200

150

100

x

y

𝑥

𝑦

y 240 181 193 155 172 110 113 75 94x 1.6 9.4 15.5 20.0 22.0 35.5 43.0 40.5 33.0

Example: The data y has been observed for various values of x, as follows:

Fit the simple linear regression model using least squares.

Answer:

n = 9 ,

�̂�=𝑆𝑥𝑦

𝑆𝑥𝑥∧�̂�=𝑦− �̂�𝑥

𝑆𝑥𝑥=∑𝑖𝑥 𝑖2−

(∑𝑖 𝑥𝑖)2

𝑛

𝑆𝑥𝑦=∑𝑖𝑥 𝑖 𝑦 𝑖−

∑𝑖𝑥 𝑖∑

𝑖𝑦 𝑖

𝑛

Want to fit

,

𝑆𝑥𝑦=26864−220.50×1333.0

9

𝑆𝑥𝑥=7053.7−220.52

9¿1651.42

¿−5794.1

⇒�̂�=𝑆𝑥𝑦

𝑆𝑥𝑥=− 5794.5

1651.45¿−3.5086

Answer:

n = 9 ,

�̂�=𝑆𝑥𝑦

𝑆𝑥𝑥∧�̂�=𝑦− �̂�𝑥

𝑆𝑥𝑥=∑𝑖𝑥 𝑖2−

(∑𝑖 𝑥𝑖)2

𝑛

𝑆𝑥𝑦=∑𝑖𝑥 𝑖 𝑦 𝑖−

∑𝑖𝑥 𝑖∑

𝑖𝑦 𝑖

𝑛

Want to fit

,

𝑆𝑥𝑦=26864−220.50×1333.0

9

𝑆𝑥𝑥=7053.7−220.52

9¿1651.42

¿−5794.1

⇒�̂�=𝑆𝑥𝑦

𝑆𝑥𝑥=− 5794.5

1651.45¿−3.5086

Now just need

�̂�=𝑦−�̂� 𝑥¿ 1333.0

9−(−3.5086)× (220.50 )

9=234.1

So the fit is approximately

Which of the following data are likely to be most appropriately modelled using a linear regression model?

1 2 3

55%

25%20%

1. Correct2. Errors change3. Not straight

1. 2.

3.

Estimating : variance of y about the fitted line Estimated error is:

, so the ordinary sample variance of the 's is

In fact, this is biased since two parameters, a and b have been estimated. The unbiased estimate is:

¿𝑆𝑦𝑦 −�̂�𝑆𝑥𝑦

𝑛−2[derivation in notes]

Quantifying the goodness of the fit

Residual sum of squares

Which of the following plots would have the greatest residual sum of squares [variance of about the fitted line]?

1 2 3

11%

72%

17%

Question from Derek Bruff

1. Pic12. Pic23. Pic correct

1. 2. 3.

Confidence interval for the slope, b

Reminder: Normal data with unknown variance, confidence interval for is:

X t snn 1

2

X t snn 1

2

to

is the estimate of , the variance of

It can be shown that , estimated by ( degrees of freedom).

b t

Snxx

2

2 b t

Snxx

2

2to

Confidence interval for b is

E.g. if you want to see if is significantly non-zero

Predictions

403020100

250

200

150

100

x

y

For given of interest, what is mean ?

Predicted mean value: .

It can be shown that

Confidence interval for mean y at given x

Extrapolation: Often not reliable

What is the error bar?