Confidence Intervals: Estimating Population Parameters with Given Levels of Certainty, Lecture notes of Statistics

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Confidence Intervals

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CONFIDENCE INTERVALS provide an interval estimate of the unknown

population parameter.

What is a confidence interval?

Statisticians have a habit of hedging their bets. They always insert qualifiers into

reports, warn about all sorts of assumptions, and never admit to anything more

extreme than probable. There's a famous saying:

"Statistics means never having to say you're certain."

Statements must be qualified, of course, because we are always dealing with

imperfect information. In particular, it is often necessary to make statements about a

population using information from a sample. No matter how carefully this sample is

selected to be a fair and unbiased representation of the population, relying on

information from a sample will always lead to some level of uncertainty.

So, a confidence interval is an interval within which we can estimate, with

some confidence , that the true population parameter will lie.

Introduction

Suppose we were interested in answering a simple research question such as:

"What is the mean number of digits that can be remembered?"

Having specified the population of people to be: “Lincoln University students”, we

take a sample of 10. The number of digits remembered for these 10 students is: 4, 4,

5, 5, 5, 6, 6, 7, 8, 9. From these results we find the estimated value of , that is x ,

to be 5.9 and s  1. 66.

But this will certainly not be a perfect estimate. It is bound to be at least either a little

too high or a little too low.

For the estimate of  to be of value, we need to have some idea of how precise it is.

That is, how close to  is the estimate likely to be?

An excellent way to specify the precision is to construct a confidence interval.

Since we know that approximately 68% of a distribution lies within 1 s.d. of the mean,

we could say that we are 68% certain that the population mean lies within an interval

of x  1s.d. That is, we could be about 68% confident that the true mean number of

digits that can be remembered lies between 5. 9  1. 66 or between 4.24 and 7.56.

And, since we know that approximately 95% of a distribution lies within 2 s.d. of the

mean, we could say that we are about 95% certain that the population mean lies

within an interval of x 2s.d. or between 5. 9  2  1. 66 i.e. between 0.92 and

10.88.

Similarly if approximately 99% of a distribution lies within 3 s.d. of the mean, we

could say that we are about 99% certain that the population mean lies within an

interval of x  3s.d. or between 5. 9  3  1. 66 i.e. between 4.24 and 7.56.

Interpretation:

A 95% confidence interval estimate means that if all possible samples are taken, 95%

of them would include the true population mean somewhere in their interval.

Or we can be 95% confident the interval contains the true population mean.

(Other confidence intervals used more frequently are 90% CI or 9 9 % CI).

If the only information given is mean and sample standard deviation or variance,

a t score is used instead of a Zscore

Use   n

s C. I . x  tn- 1 

1. From the sample, calculate x. (This may be given to you.)
2. Calculate the degrees of freedom. For a one sample mean, this is n  1.
3. Look up a t (^) score from the t- table (*see below). Note level of confidence

required and use the correct degrees of freedom ( df ).

5. Calculate the standard error of the sample statistic.

For a one sample mean,   n

s se mean .

Example: For a set of data, x  85 , s  30,andn  30, find a 95% C.I.

se ( mean ) ’ degrees of freedom  n  1   29 and 95%  upper tail = 0.025.

So from table: t score:

df t.100 t.050 t.025 t.010 t.005 t.001 t.

…..

29 1.311 1.699 2.045 2.462 2.756 3.396 3.

…

That is, t.025 = 2.045.

Hence, the C.I. = 30

=(73.7934, 96.2006)

That is, we can be 95% confident that the true population mean lies between

approximately 73.79 and 96.20.

Note that this interval is only slightly greater than the one calculated previously using

population s.d and a Zscore.

 or unknown 

2

Practise Questions – one sample

1. A machine manufactures bolts to a set length with variance of 6.25 mm. A

random sample of 20 bolts is checked and found to have a mean length of 75.2 mm. Find the 99% confidence interval for the mean length of the bolts.

2. 60 people were asked to measure their pulse rates after completing a 3 km

run. The mean was 105 beats and the standard deviation was 8 beats.

Construct a 95% confidence interval for the mean of the population of people.

3. A type of golf ball is tested, by dropping it onto a hard surface from a height of

1 metre. The height it bounces is known to be normally distributed with a

standard deviation of 3.6 cm. If a sample of 100 balls are tested and the

mean height of the bounces is 82 cm, find a. 90% b. 95% c. 99%

confidence intervals for the mean of the bounce of the golf ball.

4. A sample of stalactites (a type of rock formation) found in a glow worm cave

produced the following lengths in cm: 9.6 16.9 15.1 14.3 15.9 17.2 13.

17.1 15.4 16.2 4.5 20.3 21.2 15.

Assuming that this sample came from a normal population, calculate a 95%

confidence interval for the mean length of stalactites in the cave.

5. A doctor conducts a small survey with a random sample of his patients,

measuring their cholesterol levels.

Here is his data (the measurements are in m.mol/L):

3.6 6.9 5.1 4.2 5.5 7.2 3.0 5.8 4.9 9.9 7. 5.4 6.2 4.5 6.3 8.2 5.7 4.4 7.9 3.

Find an 80% confidence interval for the mean cholesterol level of his patients.

6. A major department store chain is interested in estimation the average amount

its credit card users spent on their first visit to the chain’s new store. Fifteen

credit cards were randomly sampled and analysed to show a mean of $50. and variance 400. Construct a 95% confident interval for the average amount

its credit card users spent on their first visit to the chain’s new store assuming

that the amount spent follows a normal distribution.

7. A race car driver tested his car for the time he takes to accelerate from 0 to 60

km/hr. In 20 such tests he obtained an average of 4.85 seconds with a

standard deviation of 1.47 seconds. What is a 95% confidence interval for

the acceleration time?

8. The actual voltages of power packs labelled as 12 volts are as follows:

11.77, 11.90, 11.64, 11.84, 12.13, 11.99, and 11.77. Calculate a 99% confidence interval for the true voltage in these packs.

When reading a question, note:

 Has the variance, standard deviation or standard error been given?

 Adjust your formula to match what has been given.

 Is the information from the population or the sample?

 Remember to use a Zscore if it is from the population and t score for a sample.

2. A political candidate finds that in a random sample of 300 constituents, 34%

support her party. Find the 95% confidence interval for the support she in fact

has.

3. Houses on a street are numbered from 1 to 627. Roimata takes a random

sample of 40 houses. She finds that in 25 of them, there are more than 3

residents. Find a 90% confidence interval for the proportion of all houses in the street having more than three residents.

4 A toy manufacturer wants to test for the proportion of faulty toys in a large

batch produced by a particular factory. He tests a random sample of 200 toys

and finds that 25 are faulty. Calculate a 94% confidence interval for the proportion of faulty toys in the complete batch.

5. In a survey carried out in Auckland, 38 people out of a random sample of 70

people said that they bought the New Zealand Herald regularly. Find a 99%

confidence interval for the proportion of people who buy the Herald in Auckland.

Answers C.I.(mean)

1 Population variance given, so use Zscore, and calculate standard dev.

6. 25 2. 5

2    99 %  Z  2. 58

 73. 75 , 76. 64  20

Hence C. I . 75. 2  2. 58  .

2 Sample standard deviation given so use t score

 102. 9 , 107. 1  60

C. I . 105  2. 009  

3. Population std.dev. given, so use Zscore.

(a) 90 %  Z  1. 645  81. 4 , 826  100

C. I . 82  1. 645  .

(b) 95 %  Z  1. 96  81. 3 , 827  100

C. I . 82  1. 96  .

(c) 99 %  Z  2. 58  81. 07 , 8293  100

C. I . 82  2. 58  .

4. Calculate sample mean and s.d.: x  15. 17 , s  4. 17

95 % anddf  13  t  2. 16  12. 76 , 17. 58  14

C. I . 15. 17  2. 16  

5. Calculate sample mean and s.d.: x  5. 75 , s  1. 768

80 % anddf  19  t  1. 328  5. 22 , 6. 28  20

C. I . 5. 75  1. 328  

That is, we can be 95% confident, credit card users spent on average between $

and $62.

 

50 5 400 20 15 14 0 025 2 1448

20 50 5 2 1448 39 42 61 58

15

x. ,s ,n df , p. t.

C.I.... ,.

        

   

That is, we can be 95%confident that the true acceleration time is between

4.16 and 5.54 seconds.

8. (^) x  11 86_. ,s_ 0 1614_._ (after entering data in calculator).

That is we can be 99% confident that the true voltage in the power packs is between 11.63 and 12.09 volts.

Answers C.I.(proportion)

1. a  0. 37 , 0. 71 

60

C. I. 0. 54 2. 58 

b.  0. 34 , 0. 46  200

CI   

c.  0. 43 , 0. 47  1000

C. I. 0. 45 1. 48 

2. n  300 , p  0. 34 , 95 %  Z  1. 96

 0. 29 , 0. 39  300

C. I. 0. 34 1. 96 

3 n  40 , 0. 625

40

p  ^90 %^ ^ Z ^1.^645

 0. 499 , 0. 751  40

C. I. 0. 625 1. 645 

4.  0. 081 , 0. 169 

200

C. I. 0. 125 1. 88 

5.  0. 389 , 0. 696 

70

70

32 10

38

C I 

 

4 85 1 47

20 19 0 025 2 093

1 47 4 85 2 093 4 16 5 54

20

x. , s.

n d. f. , p. t.

. C.I.... ,.

 

     

   

 

7 6 0 005 3 7074

0 1614 11 86 3 7074 11 63 12 09

7

n df , p. t.

. C.I.... ,.

     

   