STA 406 - Statistical Inference

Ayesha Sultan

Lecturer Virtual university of Pakistan

STA 406 - Statistical Inference

Lecture No.3

Confidence Interval Estimates• CONFIDENCE

INTERVAL for

• where:• t = Critical value from t-

distribution with n-1 degrees of freedom

• = Sample mean• s = Sample standard

deviation• n = Sample size

For very small samples (n < 30) and

is unknown.n

stx

x

Fundamental principles for using the t-distribution for confidence intervals:

1. You cannot use the t-distribution unless you assume that the population distribution of the variable is normally distributed.

2. The t-distribution, like the z-distribution, is bell-shaped and symmetric about a mean of 0.

3. The t-distribution incorporates the fact that for smaller sample sizes the distribution will be more spread out using something called degrees of freedom.

4. For every change in degrees of freedom, the t-distribution changes. The larger the sample size (n), the closer the t-distribution mimics the z-distribution in shape. We construct a confidence interval for a small sample size in the same way as we do for a large sample, except we use the t-distribution instead of the z-distribution.

Degree of FreedomThe degree of freedom for an estimate is equal to the number of values minus the number of parameters estimated en route to the estimate. For example, if there are two values (8 and 5) and we had to estimate one parameter (μ) on the way to estimating the parameter of interest (σ2). Therefore, the estimate of variance has 2 - 1 = 1 degree of freedom. Similarly, if there are 12 sampled observations then our estimate of variance would have had 11 degrees of freedom as the degrees of freedom of an estimate of variance is equal to n - 1, where n is the number of observations.

n2

i2 i=1

(X X)s =

n-1

Question 1 :You know the population mean for a certain test score. You select 10 people from the population to estimate the standard deviation. How many degrees of freedom does your estimation of the standard deviation have?

Answer :There are 10 independent pieces of information, so there are 10 degrees of freedom.

Question 2:You do not know the population mean for a different test score. You select 15 people from the population and use this sample to estimate the mean and standard deviation. How many degrees of freedom does your estimation of the standard deviation have?

Answer:The degree of freedom for an estimate is equal to the number of values minus the number of parameters estimated en route to the estimate in question. You have 15 values in your sample, and you need to estimate one parameter, the mean, in order to find the standard deviation. 15 - 1 = 14.

t distributions• Very similar to Z~N(0, 1)• Sometimes called Student’s t distribution;

Properties:

i) symmetric around 0 (like z)

ii) degrees of freedom

if > 1, E( ) = 0

if > 2, = - 2, which is always

bigger than 1.

t

-3 -2 -1 0 1 2 3

Z

0 1 2 3-1-2-3

z = x - x

x

t =

x -

s, s =

s

nx

x

x

Student’s t Distribution

-3 -2 -1 0 1 2 3

Z

t

0 1 2 3-1-2-3

n

x-x

= z

n

s - x

=t x

Student’s t Distribution

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

n

sx - x

=t

Student’s t Distribution Degrees of Freedom

s = s2

s = (X X)

n -12

i2

i=1

n

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

t7

Student’s t Distribution

n

sx - x

=t Degrees of Freedom

s = s2

s = (X X)

n -12

i2

i=1

n

df\p 0.40 0.25 0.10 0.05 0.025 0.01 0.005 0.0005 1 0.324920 1.000000 3.077684 6.313752 12.70620 31.82052 63.65674 636.61922 0.288675 0.816497 1.885618 2.919986 4.30265 6.96456 9.92484 31.59913 0.276671 0.764892 1.637744 2.353363 3.18245 4.54070 5.84091 12.92404 0.270722 0.740697 1.533206 2.131847 2.77645 3.74695 4.60409 8.61035 0.267181 0.726687 1.475884 2.015048 2.57058 3.36493 4.03214 6.8688

 6 0.264835 0.717558 1.439756 1.943180 2.44691 3.14267 3.70743 5.95887 0.263167 0.711142 1.414924 1.894579 2.36462 2.99795 3.49948 5.40798 0.261921 0.706387 1.396815 1.859548 2.30600 2.89646 3.35539 5.04139 0.260955 0.702722 1.383029 1.833113 2.26216 2.82144 3.24984 4.780910 0.260185 0.699812 1.372184 1.812461 2.22814 2.76377 3.16927 4.5869

 11 0.259556 0.697445 1.363430 1.795885 2.20099 2.71808 3.10581 4.437012 0.259033 0.695483 1.356217 1.782288 2.17881 2.68100 3.05454 4.317813 0.258591 0.693829 1.350171 1.770933 2.16037 2.65031 3.01228 4.220814 0.258213 0.692417 1.345030 1.761310 2.14479 2.62449 2.97684 4.140515 0.257885 0.691197 1.340606 1.753050 2.13145 2.60248 2.94671 4.0728

 16 0.257599 0.690132 1.336757 1.745884 2.11991 2.58349 2.92078 4.015017 0.257347 0.689195 1.333379 1.739607 2.10982 2.56693 2.89823 3.965118 0.257123 0.688364 1.330391 1.734064 2.10092 2.55238 2.87844 3.921619 0.256923 0.687621 1.327728 1.729133 2.09302 2.53948 2.86093 3.883420 0.256743 0.686954 1.325341 1.724718 2.08596 2.52798 2.84534 3.8495

 21 0.256580 0.686352 1.323188 1.720743 2.07961 2.51765 2.83136 3.819322 0.256432 0.685805 1.321237 1.717144 2.07387 2.50832 2.81876 3.792123 0.256297 0.685306 1.319460 1.713872 2.06866 2.49987 2.80734 3.767624 0.256173 0.684850 1.317836 1.710882 2.06390 2.49216 2.79694 3.745425 0.256060 0.684430 1.316345 1.708141 2.05954 2.48511 2.78744 3.7251

Degrees of Freedom1 3.0777 6.314 12.706 31.821 63.6572 1.8856 2.9200 4.3027 6.9645 9.9250. . . . . .. . . . . .

10 1.3722 1.8125 2.2281 2.7638 3.1693. . . . . .. . . . . .

100 1.2901 1.6604 1.9840 2.3642 2.62591.282 1.6449 1.9600 2.3263 2.5758

0.80 0.90 0.95 0.98 0.99

t-Table: text- inside back cover

• 90% confidence interval; df = n-1 = 10

118125.1 :interval confidence%90

sx

0 1.8125

P(t > 1.8125) = .05

-1.8125.05.05

.90

P(t < -1.8125) = .05

Comparing t and z Critical Values

Confidence level n = 30z = 1.645 90% t = 1.6991z = 1.96 95% t = 2.0452z = 2.33 98% t = 2.4620z = 2.58 99% t = 2.7564

Example An investor is trying to estimate the return on investment in companies that won quality awards last year. A random sample of 25 such companies is selected, and the return on investment is recorded for each company. The data for the 25

companies have

Construct a 95% confidence interval for the mean return.

18.875.14 sx

14.75 8.18

degrees of freedom 25 1 24

x s

from t-table, t 2.0211

8.1814.75 2.064

25

14.75 3.376 11.37,18.12

We are 95% confident that the interval

(11.37,18.12) contains the population mean

return on investment for companies that win

qualit

sx t

n

y awards.

1.. nfdn

stx

ExampleCardiac deaths increase after heavy snowfalls, a study was conducted to measure the cardiac deaths of shoveling snow by hand. The maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation were 175 and 15 respectively.

Find a 90% CI for the population mean maximum heart rate for those who shovel snow.

Solution 1.. nfdn

stx

shovelers snowfor rateheart maximum

mean thecontains 183.70) (166.30,

interval that theconfident 90% are We

)70.183,30.166(

70.817510

158331.1175

1.8331 ttable,- t theFrom

1015,175

nsx

Example

The masses, in grams, of twelve ball bearings taken at random from a batch are 31.4, 33.1, 35.9, 34.7, 33.4, 34.5, 35.0, 32.5, 36.9, 36.4, 35.8 and 33.2. Calculate a 90% confidence interval for the mean mass of the population, supposed normal, from which these masses were drawn.

Confidence Interval Estimates for 1 - 2

STANDARD DEVIATIONS UNKNOWN

AND 12 = 2

2

where: = Pooled standard deviation

t/2 = critical value from t-distribution for desired confidence level and degrees of freedom equal to n1 + n2 -2

212/21

11)(

nnstxx p

2

)1()1(

21

222

211

nn

snsnsp

Example

a) Give two random samples of sizes 1 29& 16n n from two independent normal

populations, with

1 2 1 264, 59, 6 5,x x s and s find a 95% confidence interval for 1 2 assuming that

1 2 .

b) A sample from a normal population with unknown variance consists of the observations 34, 25, 43, 37, 45. A sample from a second normal population with the same unknown variance as the first consists of the observations 20, 31, 23, 35, 41, 29, 39. Find a 95% confidence interval for 1 2

Confidence Interval Estimates for 1 - 2

STANDARD DEVIATIONS UNKNOWN

AND 12 2

2

where: t/2 = critical value from t-distribution for desired confidence level

and degrees of freedom equal to:

2

22

1

21

2/21 )(n

s

n

stxx

)1

)/(

1

)/((

)//(

2

22

22

1

21

21

2221

21

n

ns

n

ns

nsns

Confidence Interval Estimate for paired samples

Paired samples are samples that selected such that each data value from one sample is related (or matched) with a corresponding data value from the second sample. The sample values from one population have the potential to influence the probability that values will be selected from the second population.

Confidence Interval Estimate for paired samples

PAIRED CONFIDENCE INTERVAL ESTIMATE

n

std d

2/

Confidence Interval Estimate for paired samples

PAIRED DIFFERENCE

where: d = Paired difference

x1 and x2 = Values from sample 1 and 2, respectively

21 xxd

Confidence Interval Estimate for paired samples

MEAN PAIRED DIFFERENCE

where: di = ith paired difference

n = Number of paired differences

1

n

ii

dd

n

STANDARD DEVIATION FOR PAIRED DIFFERENCES

where: di = ith paired difference

= Mean paired difference

1

)(1

2

n

dds

n

ii

d

d

ExampleA nutrition scientist is assessing a weight-loss programme toevaluate its effectiveness. Ten people were randomly selected. Both the initial weight and the final weight after 20 weeks on the programme was recorded as:

Subject Initial Weight Final Weight

1 180 165

2 142 138

3 126 128

4 138 136

5 175 170

6 205 197

7 116 115

8 142 128

9 157 144

10 136 130

Find the 95% confidence interval for the mean difference between Initial weight and Final weight. Assume that the mean differences are approximately normally distributed.

ExampleTwenty-two students were randomly selected from a population of 1000 students. All of the students were given a standardized English test and a standardized Math test. Find the 90% confidence interval for the mean difference between student scores on the Math and English tests. Assume that the mean differences are approximately normally distributed.

Student English(x) Math(Y)1 95 902 89 853 76 734 92 905 91 906 53 537 67 688 88 909 75 78

10 85 8911 90 9512 85 8313 87 8314 85 8315 85 8216 68 6517 81 7918 84 8319 71 6020 46 4721 75 7722 80 83

Test results are summarized below: