A Confidence Interval Provides Additional Information About Variability

7/29/2019 A Confidence Interval Provides Additional Information About Variability

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A Confidence interval provides additional information about variability. It consists of an

interval of numbers along with a probability that the interval contains the unknown parameter

A point estimate of a parameter is a single sample value of a statistic that estimates the value

of the parameter.

The level of confidence in a confidence interval is a probability that represents the

percentage of intervals that will contain if a large number of repeated samples are obtained.

The level of confidence is denotedFor example, a 95% level of confidence would mean that

if 100 confidence intervals were constructed, each based on a different sample from the same

population, we would expect 95 of the intervals to contain the population mean.

The general formula for all confidence intervals is: Point Estimate (Critical Value)(Standard Error)

The construction of a confidence interval for the population mean depends upon three

factors-:

The point estimate of the population The level of confidence The standard deviation of the sample mean

Confidence Interval for

( Known)

95% of all sample means are in the interval

where X is the point estimate

Z(1.96) is the normal distribution critical value for a probability of/2 in each tail


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Levels Of Confidence

Sampling Distribution of the Mean

Ques 1 A sample of 11 circuits from a large normal population has a mean resistance of 2.20

ohms. We know from past testing that the population standard deviation is 0.35 ohms.

Determine a 95% confidence interval for the true mean resistance of the population

ConfidenceLevel

Confidence

Coefficient, Z value

1.28

1.645

1.96

2.332.58

3.08

3.27

0.80

0.90

0.95

0.980.99

0.998

0.999

80%

90%

95%

98%99%

99.8%

99.9%

1

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2.40681.9932

0.20682.20

)11(0.35/1.962.20n

ZX


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Interpretation

We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms

Although the true mean may or may not be in this interval, 95% of intervals formed in this

manner will contain the true mean

Confidence Interval for

( Unknown)

Note:

If the population standard deviation is unknown, we can substitute the sample

standard deviation, S This introduces extra uncertainty, since S is variable from

sample to sample.So we use the t distribution instead of the normal distribution

The t value depends on degrees of freedom (d.f.)

Number of observations that are free to vary after sample mean has been calculated

d.f. = n1

Example 2

A random sample of n = 25 taken from a normal population has X = 50 and S = 8. Form a95% confidence interval for

n

StX

1-n

2.0639t0.025,241n,/2 t


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d.f. = n1 = 24, so

The confidence interval is

46.698 53.302

Confidence Intervals for the Population Proportion,

An interval estimate for the population proportion ( ) can be calculated by adding and

subtracting an allowance for uncertainty to the sample proportion ( p )

SD of Interval

Sample Size

Upper and lower confidence limits for the population proportion are calculated with

the formula-:

where

Z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size

n

)(1p

n

p)p(1

25

8(2.0639)50

n

S1-n/2, tX

n

p)p(1Zp


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Example 4 A random sample of 100 people shows that 25 are left-handed. Form a 95%

confidence interval for the true proportion of left-handers

Interpretation

We are 95% confident that the true percentage of left-handers in the population is

between 16.51% and 33.49%.

Sampling Error is the amount of imprecision in the estimate of the population parameter

To determine the required sample size for the mean, you must know:

The desired level of confidence (1 - ), which determines the critical Z value The acceptable sampling error, e The standard deviation,

s

If unknown, can be estimated when using the required sample size formula Use a value for that is expected to be at least as large as the true

/1000.25(0.75)1.9625/100

p)/np(1Zp

0.33490.1651

(0.0433)1.960.25

2

22

e

Zn

219.195

(45)(1.645)2

22

2

22

e

Zn


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Select a pilot sample and estimate with the sample standard deviation, S

To determine the required sample size for the proportion, you must know: The desired level of confidence (1 - ), which determines the critical Z value The acceptable sampling error, e The true proportion of successes,

can be estimated with a pilot sample, if necessary (or conservativelyuse = 0.5)

Example 5

How large a sample would be necessary to estimate the true proportion defective in a large

population within 3%, with 95% confidence?

(Assume a pilot sample yields p = 0.12)

For 95% confidence, use Z = 1.96

e = 0.03

p = 0.12, so use this to estimate

2

2

e

)(1Z

n

450.74(0.03)

0.12)(0.12)(1(1.96)e

)(1Zn2

2

2

2


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Fundamentals of Hypothesis Testing: One-Sample Tests

A hypothesis is a claim (assumption) about a population parameter.

Population mean

Example: The mean monthly cell phone bill of this city is = $42

Population proportion

Example: The proportion of adults in this city with cell phones is = 0.68

Null Hypothesis states the claim or assertion to be tested

Example: The average number of TV sets in U.S. Homes is equal to three Ho: = 3

Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until

proven guilty

May or may not be rejectedAlternative hypothesis HiIs the opposite of the null hypothesis

e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: 3 )

May or may not be proven Is generally the hypothesis that the researcher is trying to prove

Level of Significance,

Defines the unlikely values of the sample statistic if the nullhypothesis is true

Defines rejection region of the sampling distribution

Is designated by , (level of significance).Typical values are 0.01, 0.05, or0.10


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Type I Error

Reject a true null hypothesis Considered a serious type of error

The probability of Type I Error is ,called level of significance of the test Type II Error

Does Not reject a false null hypothesisThe probability of Type II Error is

Hypothesis Test for Mean

ActualSituation

Decision

Do NotReject

H0

No error

(1-

)

Type II Error

( )

RejectH0

Type I Error( )

H0

FalseH0

True

No Error( 1 - )

Known Unknown

Hypothesis

Tests for

(Z test) (t test)


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When SD Is Known (Z test )

For a two-tail test for the mean, known: Convert sample statistic ( ) to test statistic (Z statistic ) Determine the critical Z values for a specified

level of significance from a table or computer Decision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do

not reject H0

Example

Test the claim that the true mean # of TV sets in US homes is equal to 3.

(Assume = 0.8)

H0: = 3 H1: 3 (This is a two-tail test)

Suppose that = 0.05 and n = 100 are chosen for this test

is known so this is a Z test.

For = 0.05 the critical Z values are 1.96

Suppose the sample results are

n = 100, X = 2.84 ( = 0.8 is assumed known)

So the test statistic is:

Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficientevidence that the mean number of TVs in US homes is not equal to 3

n

X

Z

2.0.08

.16

100

0.8

32.84

n

XZ


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p-Value Approach to Testing

p-value: Probability of obtaining a test statistic more extreme ( or ) than the observedsample value given H0 is true

Also called observed level of significance. It is the smallest value of forwhich H0 can be rejected

1)Convert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic )

2)Obtain the p-value from a table

3)Compare the p-value with If p-value < , reject H0 If p-value , do not reject H0

Example

How likely is it to see a sample mean of 2.84 (or something further from the mean, in

either direction) if the true mean is = 3.0?

X = 2.84 is translated to a Z score of Z = -2.0

p-value

= 0.0228 + 0.0228 = 0.0456

Compare the p-value with If p-value < , reject H0 If p-value , do not reject H0

Here: p-value = 0.0456

= 0.05Since 0.0456 < 0.05, we reject the null hypothesis

Connecting the above to confidence interval

For X = 2.84, = 0.8 and n = 100, the 95% confidence interval is:

2.6832 2.9968 Since interval does not contain hypothesized mean 3.0 we

reject the null hypothesis at a=0.05

0.02282.0)P(Z

0.02282.0)P(Z

100

0.8(1.96)2.84to

100

0.8(1.96)-2.84


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One Tail Tests

H1: < 3

H0: 3 This is a lower/left-tail test since the alternative hypothesis is focused on the

lower/left tail below the mean of 3

H1: > 3

H0: 3 This is an upper/right-tail test since the alternative hypothesis is focused on the

upper/right tail above the mean of 3

Example Upper/Right-Tail Z Test

for Mean ( Known)

A phone industry manager thinks that customer monthly cell phone bills have

increased, and now average over $52 per month. The company wishes to test thisclaim. (Assume = 10 is known)

Suppose that = 0.10 is chosen for this test thus Rejection region would be +1.28= 0.8997

from table

Suppose a sample is taken with the following results: n = 64, X = 53.1 (=10 was assumed

known) Then the test statistic is:

Do not reject H0since Z = 0.88 1.28

i.e.: there is not sufficient evidence that the mean bill is over $ 52

Calculating P value

H0: 52 the average is not over $52 per month

H : > 52 the average is greater than $52 per month

0.88

64

10

5253.1

n

XZ

0.1894

0.810610.88)P(Z

6410/

52.053.1ZP

53.1)XP(


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Do not reject H0 since p-value = 0.1894 > = 0.10

t Test of Hypothesis for the Mean ( Unknown)

Example -The average cost of a hotel room in New York is said to be $168 per night.

A random sample of 25 hotels resulted in X = $172.50 and

S = $15.40. Test at the H0: = 168

= 0.05 level. H1: 168

a = 0.05

n= 25

is unknown, so use a t statistic

Critical Value: t24 = 2.0639

Do not reject H0: not sufficient evidence that true mean cost is different than

$168

Connecting to confidence interval

For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is:172.5 - (2.0639) 15.4/ 25 to 172.5 + (2.0639) 15.4/ 25

166.14 178.86

Since this interval contains the Hypothesized mean (168), we do not reject the nullhypothesis at = 0.05

Proportions

Sample proportion in the success category is denoted by p

n

SXt 1-n

1.46

25

15.40

168172.50

n

S

Xt

1n

sizesample

sampleinsuccessesofnumber

n

Xp


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When both n and n(1-) are at least 5, pcan be approximated by a normal distribution

with mean and standard deviation

The sampling distribution of p is approximately normal, so the test statistic is a Zvalue:

Example

A marketing company claims that it receives 8% responses from its mailing. To test this

claim, a random sample of 500 were surveyed with 25 responses. Test at the = 0.05

significance level.

Check:

n = (500)(.08) = 40 H0: = 0.08

n(1-) = (500)(.92) = 460 Hi:Y=/o.o8

There is sufficient evidence to reject the companys claim of 8% response rate.

p-Value Solution

Reject Ho since p-value=0.0136


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University of theFraser Valley

Assignment Chapter 8&9

Presented To Prof Gagan Sharma

Presented By Shrey Budhiraja

200106769

A Confidence Interval Provides Additional Information About Variability

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