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ES Chapter 18 & 20: Inferences Involving One Population

0 t

Student’s t, df = 5

Student’s t, df = 15Student’s t, df = 25

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ES Chapter Goals

• Learned about confidence intervals

• Assumed was known

• Consider inference about when is unknown

• Consider inference about p, the probability of success

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• Inferences about are based on the sample mean x

• If the sample size is large or the sample population is normal: has a standard normal distribution

)//()(* nxz

Inference About mean ( unknown)

• If is unknown, use s as a point estimate for

ns /• Estimated standard error of the mean:

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1. When s is used as an estimate for , the test statistic has two sources of variation: sx and

4. The population standard deviation, , is almost never known in real-world problems

The standard error will almost always be estimated using

Almost all real-world inference about the population mean will be completed using the Student’s t-statistic

ns

Student’s t-Statistic

3. Assumption: samples are taken from normal populations

2. The resulting test statistic:

nsxt Known as the Student’s t-statistic

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ES Properties of the t-Distribution (df>2)

3. t is distributed so as to form a family of distributions, a separate distribution for each different number of degrees of freedom

)1df(

1. t is distributed with a mean of 0

2. t is distributed symmetrically about its mean

4. The t-distribution approaches the normal distribution as the number of degrees of freedom increases

5. t is distributed with a variance greater than 1, but as the degrees of freedom increase, the variance approaches 1

6. t is distributed so as to be less peaked at the mean and thicker at the tails than the normal distribution

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0 t

Student’s t, df = 5Student’s t, df = 15

Normal distribution

Degrees of Freedom, df: A parameter that identifies each different distribution of Student’s t-distribution. For the methods presented in this chapter, the value of df will be the sample size minus 1, df = n 1.

Student’s t-Distributions

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1. The number of degrees of freedom associated with s2 is the divisor (n 1) used to define the sample variance s2

Thus: df = n 1

Notes

2. The number of degrees of freedom is the number of unrelated deviations available for use in estimating 2

3. Table for Student’s t-distribution (Table C) is a table of critical values. Left column = df. When df > 100, critical values of the t-distribution are the same as the corresponding critical values of the standard normal distribution.

4. Notation: t(df, )

Read as: t of df,

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t0 )df, (t

t-Distribution Showing t(df, )

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ES ExampleExample: Find the value of t(12, 0.025)

Amount of in one-taildf 0.025

12 2.18

Portion ofTable 6

...

. . . . . .

0- t (12, 0.025) t18.218.2

025.0025.0

t (12, 0.025)

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1. If the df is not listed in the left-hand column of Table C, use the next smaller value of df that is listed

Notes

t

cumulative probability

2. Most computer software packages will calculate either the area related to a specified t-value or the t-value that bounds a specified area

3. The cumulative distribution function (CDF) is often used to find

area from to t

4. If the area from to t is known and the value of t is wanted, then the inverse cumulative distribution function (INVCDF) is used

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Confidence Interval Procedure:

1. Procedure for constructing confidence intervals similar to that used when is known

The assumption for inferences about mean when is unknown: The sampled population is normally distributed

The Assumption...

2. Use t in place of z, use s in place of

3. The formula for the confidence interval for is:

1df wheret(df, to nn

sxn

sx t(df,

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Example: A study is conducted to learn how long it takes the typical tax payer to complete their federal income tax return. A random sample of 17 income tax filers showed a mean time (in hours) of 7.8 and a standard deviation of 2.3. Find a 95% confidence interval for the true mean time required to complete a federal income tax return. Assume the time to complete the return is normally distributed.

Solution:1. Parameter of Interest

The mean time required to complete a federal income tax return

Example

2. Confidence Interval Criteriaa. Assumptions: Sampled population assumed normal, unknownb. Test statistic: t will be usedc. Confidence level: = 0.95

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3.2 and ,8.7 ,17 sxn3. The Sample Evidence:

Solution Continued

5. The Results:

6.62 to 8.98 is the 95% confidence interval for

4. The Confidence Intervala. Confidence coefficients:b. Maximum error:

c. Confidence limits:

18.1)5578.0)(12.2(173.2)12.2(

nsE

98.8 to62.618.18.7 to18.18.7

to

ExEx

t(df, /2) = t(16, 0.025) = 2.12

t(16, 0.025)

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ES Inferences About theProbability of Success

• Possibly the most common inference of all

• Many examples of situations in which we are concerned about something either happening or not happening

• Two possible outcomes, and multiple independent trials

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1. p: the binomial parameter, the probability of success on asingle trial

Background

2. : the observed or sample binomial probability'p

nxp ' x represents the number of successes that

occur in a sample consisting of n trials

pqnpqnp 1 where, , 3. For the binomial random variable x:

4. The distribution of x is approximately normal if n is larger than 20 and if np and nq are both larger than 5

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1. a mean equal to p,2. a standard error equal to , and3. an approximately normal distribution if n is sufficiently large

'p

'p npq /)(

Sampling Distribution of p': If a sample of size n is randomly selected from a large population with p = P(success), then the sampling distribution of p' has:

Sampling Distribution of p'

In practice, use of the following guidelines will ensure normality:1. The sample size is greater than 202. The sample consists of less than 10% of the population3. The products np and nq are both larger than 5

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The assumptions for inferences about the binomial parameter p: The n random observations forming the sample are selected independently from a population that is not changing during the sampling

The Assumptions...

Confidence Interval Procedure:The unbiased sample statistic p' is used to estimate the population proportion p

'1' and /' where pqnxp

The formula for the confidence interval for p is:

nqpp

nqpz(p ''' to''' z(

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Example: A recent survey of 300 randomly selected fourth graders showed 210 participate in at least one organized sport during one calendar year. Find a 95% confidence interval for the proportion of fourth graders who participate in an organized sport during the year.

Solution:1. Describe the population parameter of concern

The parameter of interest is the proportion of fourth graders who participate in an organized sport during the year

Example

2. Specify the confidence interval criteriaa. Check the assumptionsThe sample was randomly selectedEach subject’s response was independent

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590)300/90(300'5210)300/210(300'

20300

nqnpn

b. Identify the probability distribution z is the test statisticp' is approximately normal

c. Determine the level of confidence : 0.95

Solution Continued

70.0300/210/' nxp

3. Collect and present sample evidenceSample information: n = 300, and x = 210The point estimate:

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4. Determine the confidence intervala. Determine the confidence coefficients:Using Table A: z( /2) = z(0.025) = 1.96

Solution Continued

7519.0 to6481.00519.070.0 to0519.070.0

' to'

EpEpc. Find the lower and upper confidence limits:

0519.0)0265.0)(96.1(0007.0)96.1(300

)30.0)(70.0(96.1''

nqpE

b. The maximum error of estimate:

z( /2)

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d. The Results0.6481 to 0.7519 is a 95% confidence interval for the true proportion of fourth graders who participate in an organized sport during the year

Solution Continued

E: maximum error of estimate: confidence levelp*: provisional value of p (q* = 1 p*)If no provisional values for p and q are given use p* = q* = 0.5(Always round up)

2

2 **][E

qpnSample Size Determination: z( /2)

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Example: Determine the sample size necessary to estimate the true proportion of laboratory mice with a certain genetic

defect. We would like the estimate to be within 0.015 with 95% confidence.

Example

Solution:1. Level of confidence: = 0.95, z(/2) = z(0.025) = 1.962. Desired maximum error is E = 0.015.3. No estimate of p given, use p* = q* = 0.5

4. Use the formula for n:

4269 44.4268000225.09604.0

)015.0()5.0()5.0()96.1(**][

2

2

2

2

n

Eqpn

z( /2)

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211 210.75 000225.00474.0

)015.0()9875.0()0125.0()96.1(**][

2

2

2

2

n

Eqpn

Note: Suppose we know the genetic defect occurs in approximately 1 of 80 animals

Use p* = 1/80 = 0.125:

Note

As illustrated here, it is an advantage to have some indication of the value expected for p, especially as p becomes increasingly further from 0.5

z( /2)