Chapter 3: The Normal Distributions

Transcript Chapter 3: The Normal Distributions

Chapter 18: Inference about One Population Mean

STAT 1450

18.0 Inference about One Population Mean

Connecting Chapter 18 to our Current Knowledge of Statistics

▸ We know the basics of confidence interval estimation (Chapter 14) and tests of significance (Chapter 15). Nuances that we should be aware of were also presented (Chapter 16).

18.0 Inference about One Population Mean

Parameters and their Point Estimates

▸ Measure Mean Standard Deviation Proportion of Successes Sample Statistic and Point Estimate s 𝑝 𝑥 Population Parameter μ σ 𝑝 In the coming chapters, we will either find confidence intervals for the population parameters, or, conduct tests of significance regarding their hypothesized values . ▸ In either case, the point estimates will help us in our endeavors.

18.0 Inference about One Population Mean

Inference when σ is unknown

▸ It is unlikely that the population standard deviation

will be known and the population mean

will not be known. ▸ Chapter 14 taught us that 𝑥 is the best point estimate of µ. Similarly,

can estimate

18.0 Inference about One Population Mean

Inference when σ is unknown

▸ In Chapter 11 when σ was known we used  𝑧 = 𝑥−𝜇 𝜎 𝑛 This statistic follows a Standard Normal Z-distribution ▸ When σ is not known we can use s instead:  𝑡 = 𝑥−𝜇 𝑠 𝑛 This statistic is not quite Normal. It follows a

-distribution.

18.1 Conditions for Inference about a Population Mean

Conditions for Inference about a Mean

▸ The conditions for inference about a mean are listed on page 437 of the text. Random sample:  Do we have a random sample?

 If not, is the sample representative of the population?

 If not a representative sample, was it a randomized experiment?

18.1 Conditions for Inference about a Population Mean

Conditions for Inference about a Mean

 Large enough population : sample ratio: Is the population of interest ≥ 20 times ‘n’?

 The population is from a Normal Distribution.

If the population is not from a Normal Distribution, then the sample size must be “large enough” with a shape similar to the Normal Distribution; then we apply the Central Limit Theorem.

18.1 Conditions for Inference about a Population Mean

Standard Error

▸ When the standard deviation of a statistic is estimated from data, the result is called the standard error of the statistic. ▸ The standard error of the sample mean is 𝑠 𝑛 .

▸ Now the sample standard deviation will replace σ.

18.1 Conditions for Inference about a Population Mean

Example: Standard Error

▸ A random sample of 49 students reported receiving an average of 7.2 hours of sleep nightly with a standard deviation of 1.74. What is the standard deviation of the mean?

18.1 Conditions for Inference about a Population Mean

Example: Standard Error

▸ A random sample of 49 students reported receiving an average of 7.2 hours of sleep nightly with a standard deviation of 1.74. What is the standard deviation of the mean?

 (sample) standard deviation of 1.74

 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 (𝑠. 𝑒. ) = 1.74

49 = 0.2486

18.2 The t Distributions

The t-distribution

▸ Draw an SRS of size

from a large population that has the Normal distribution with mean

and standard deviation

. The

one-sample t statistic

𝑡 = 𝑥 − 𝜇 𝑠 𝑛 has the

t distribution

with

– 1 degrees of freedom.

18.1 Conditions for Inference about a Population Mean

Standard Error

▸ Now the sample standard deviation will replace s; allowing us to use the one-sample

statistic for confidence intervals and tests of significance. ▸ As mentioned earlier, the t distribution is “not quite Normal.”

18.2 The t Distributions

The t Distributions

▸ Here is a plot of two

distributions (dashed) and the standard Normal distribution (solid):

18.2 The t Distributions

T-distribution Compared to the Z-Distribution

Similarities Symmetric about 0 Single-peaked & Bell shaped Differences T has thicker tails Varies based upon ‘degrees of freedom’

18.2 The t Distributions

T-distribution Compared to the Z-Distribution

▸

Poll:

The t 2 curve has thicker dashes. The t 9 curve has smaller dashes. Z is the solid curve. What would you anticipate happening to t df as the degrees of freedom (df) increase?

a) t df will not be affected b) t df will approach Z c) t df will become further from Z.

18.2 The t Distributions

T-distribution Compared to the Z-Distribution

▸

Poll:

The t 2 curve has thicker dashes. The t 9 curve has smaller dashes. Z is the solid curve. What would you anticipate happening to t df as the degrees of freedom (df) increase?

a) t df will not be affected b) t df will approach Z c) t df will become further from Z.

18.2 The t Distributions

Using Table C

▸ “What is happening to these values as we increase the ‘df’?” 2/3. If the bottom row contains z*, and t-critical values increase as we ‘move up the chart,’ should we expect intervals based upon t to be larger or smaller than those based upon z?

Suppose df=37, then use row for df=30. Walk through this example.

18.2 The t Distributions

Using Table C Notes:

1. The t-distribution

critical values decrease

as the

degrees of freedom (df) increase

. 2. The final row includes 1.645, 1.96, & 2.576. These are “common confidence levels ” & z*. 3. Confidence intervals based upon “t” will be

slightly

wider than those based upon “z.” 4. Be conservative. When the exact df is not listed, “round down” and use the closest df that does not exceed the df that is desired .

18.3 The One-sample t Confidence Interval

The One-sample t Confidence Interval

▸ Draw an SRS of size

from a large population having unknown mean

. ▸ A level C confidence interval for

is ▸ 𝑥 ± 𝑡 ∗ 𝑠 𝑛 where

* is the critical value for the

(

– 1) density curve with area C between ‒

* and

*. This interval is exact when the population distribution is Normal and is approximately correct for large

in other cases.

18.3 One-Sample t Confidence Intervals

The One-sample t Confidence Interval

▸ The one-sample t confidence interval is used to estimate means. ▸ Its form is similar to previous forms of confidence intervals: estimate ± margin of error Introduction of Confidence Intervals 𝑥 ± (1.96 or another z-score) 𝜎 𝑛 General Form (when σ is known) 𝑥 ± 𝑡 ∗ 𝑠 𝑛 Now ( s is unknown).

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels

▸ Hemoglobin (Hb) levels are normally distributed, and should neither be too large, nor too small. A random sample of 11 boys from an underserved country had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Compute a 90% confidence interval for the average hemoglobin level for boys from this particular country.

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?) .

Yes. Stated as a random sample.

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?) .

Yes. Stated as a random sample.

 Large enough population: sample ratio? Yes. .

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?) .

Yes. Stated as a random sample.

 Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220.

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?) .

Yes. Stated as a random sample.

 Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220.

 t-distribution?

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?) .

Yes. Stated as a random sample.

 Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220.

 t-distribution? Yes. n = 11 < 40 …

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components

 Do we have an SRS?

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Yes. Stated as a random sample.

 Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220.

 t-distribution? Yes. n = 11 < 40 … but data is approximately Normal, so we can use t-distribution.

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

df =(n-1)=10 t*(90%, 10) = 1.812

90% df=11-1=10

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

4. Interval.

11.3 ± 1.812

1.5

11 =

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

4. Interval.

11.3 ± 1.812

1.5

11 = 11.3 ± .82

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

4. Interval.

11.3 ± 1.812

1.5

11 = 11.3 ± .82 = (10.48,12.12)

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

Interpret

* the interval.

4. Interval.

11.3 ± 1.812

1.5

11 = 11.3 ± .82 = (10.48,12.12)

18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 2. Components.

= 11.3, s = 1.5, n = 11

3. Select t*.

df =(n-1)=10 t*(90%, 10) = 1.812

Steps for Success Constructing Confidence Intervals for

(

unknown).

Confirm that the 3 key

conditions

are satisfied

(SRS?, N:n?, t-distribution?).

Identify the 3 key

components

of the confidence interval

(mean, s.d., n

Select

4. Construct the confidence

interval

Interpret

* the interval.

4. Interval.

11.3 ± 1.812

1.5

11 = 11.3 ± .82 = (10.48,12.12)

5. Interpret.

We are 90% confident that the mean hemoglobin level for boys from this country is between 10.48 g/dL and 12.12 g/dL

18.3 One-Sample t Confidence Intervals

Technology Tips – Computing Confidence Intervals (

unknown)

Technology Tips – Computing Confidence Intervals (s unknown) ▸

TI-83/84: STAT



TESTS



TInterval



Enter

 

Select

Stats.

Enter s, , 𝑥 n, and the confidence level.

Select

Calculate

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data ▸ are stored.)

JMP:

Enter the data.

Analyze



Distribution.

 

“

Click-and-Drag

” (the appropriate variable)

into the

‘Y, Columns’

box.

Click

OK.

Click

on the red upside-down triangle next to the title of the variable from the

‘Y,Columns’

box.

Proceed to ‘

Confidence Interval

’ -> 

Select the appropriate confidence level.

18.3 One-Sample t Confidence Intervals

Technology Tips – Computing 90% Confidence Intervals (

𝝈

unknown)

▸ TI-83/84  

STAT >> TESTS >> TInterval >> Enter

Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.

(for this example) 

Inpt >> STATS

  : 11.3

>> s: 1.5

C-Level : 90



Calculate (ENTER) >>

n :

18.3 One-Sample t Confidence Intervals

Technology Tips – Computing 90% Confidence Intervals (

𝝈

unknown)

▸ TI-83/84  

STAT >> TESTS >> TInterval >> Enter

Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.

(for this example) 

Inpt >> STATS

  : 11.3

>> s: C-Level : 90 1.5

n :



Calculate (ENTER) (10.48, 12.12)

18.3 One-Sample t Confidence Intervals

Technology Tips – Computing 90% Confidence Intervals (

𝝈

unknown)

▸ TI-83/84  

STAT >> TESTS >> TInterval >> Enter

Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.

(for this example) 

Inpt >> STATS

  : 11.3

>> s: C-Level : 90 1.5

n :



Calculate (ENTER) (10.728, 11.88)

18.3 One-Sample t Confidence Intervals

Technology Tips – Computing 90% Confidence Intervals (

𝝈

unknown)

▸ TI-83/84  

STAT >> TESTS >> TInterval >> Enter

Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.

(for this example) 

Inpt >> STATS

  : 11.3

>> s: C-Level : 90 1.5 >>

n :

33 Larger n, narrower interval.



Calculate (ENTER) (10.858, 11.742)

18.4 One-Sample t Test

The One-sample t Test of Significance

▸ Draw an SRS of size n from a large population that has the Normal distribution with mean μ and standard deviation σ. The one-sample t statistic 𝑡 = 𝑥 − 𝜇 𝑠 𝑛 has the t distribution with n – 1 degrees of freedom.

The One-sample t Test of Significance

▸ To test the hypothesis 𝐻 𝑜 : 𝜇 = 𝜇 𝑜 , compute the one-sample t statistic 𝑡 = 𝑠 𝑥 − 𝜇 𝑜 𝑛 18.4 One-Sample t Test

▸

The One-sample t Test of Significance

The

-value for a test of

H 0

against  𝐻 𝑎 : 𝜇 < 𝜇 0 is 𝑃 𝑇 ≤ 𝑡 .

18.4 One-Sample t Test  𝐻 𝑎 : 𝜇 > 𝜇 0 is 𝑃(𝑇 ≥ 𝑡) .

 𝐻 𝑎 : 𝜇 = 𝜇 0 is 2𝑃(𝑇 ≥ 𝑡 ) .

 These

-values are exact if the population distribution is Normal and are approximately correct for large

in other cases.

18.4 One-Sample t Test

Example: Hemoglobin levels

▸ Recall from earlier that Hb levels are normally distributed. Our original example featured a random sample of 11 boys from an underserved country that had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Is there significant evidence, at the .05 level of significance that the average Hb level for boys from this country is below 12, which results in ___________?

18.4 One-Sample t Test

Example: Hemoglobin levels

15.4 Tests for a Population Mean

Steps for Success – Conducting Tests of Significance Steps for Success Conducting Tests of Significance

1. Set up your

Hypotheses

2. Check your

Conditions

. 3. Compute the

Test Statistic

. 4. Compute the

P-Value

5. Make a

Decision

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

c) State the null (H 0 ) and alternative (H A ) hypotheses.

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

c) State the null (H 0 ) and alternative (H A ) hypotheses.

H 0 : µ = 12 H a :µ < 12

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

c) State the null (H 0 ) and alternative (H A ) hypotheses.

d) Specify the level of significance. α =.05

H 0 : µ = 12 H a :µ < 12

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

c) State the null (H 0 ) and alternative (H A ) hypotheses.

d) Specify the level of significance. α =.05

e) Determine the type of test.

Left-tailed H 0 : µ = 12 Right-tailed H a :µ < 12 Two-Tailed

18.4 One-Sample t Test

Example: Hemoglobin levels State:

Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)?

Plan:

a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

b) List all given information from the data collected.

n=11, 𝑥 = 11.3, sd=1.5

c) State the null (H 0 ) and alternative (H A ) hypotheses.

d) Specify the level of significance. α =.05

e) Determine the type of test.

Left-tailed H 0 : µ = 12 Right-tailed H a :µ < 12 Two-Tailed

Example: Hemoglobin levels Plan:

f) Sketch the region(s) of “extremely unlikely” test statistics.

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

 Large enough population: sample ratio?

Yes.

The number of boys is arbitrarily large; therefore, N > 20*11 = 220.

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

 Large enough population: sample ratio?

Yes.

The number of boys is arbitrarily large; therefore, N > 20*11 = 220.

 Large enough sample; Normal or t-distribution?

Yes.

n = 11 < 40. But data is approximately Normal, so we can use t-distribution.

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝑠 𝑛 = 18.4 One-Sample t Test

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝑠 𝑛 = 11.3 − 12 = −1.548

1.5

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) c) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝑠 𝑛 = 11.3 − 12 = −1.548

1.5

11 Determine (or approximate) the P-Value.

1.548

 -1.372 > -1.548 > -1.812

-value

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) c) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝑠 𝑛 = 11.3 − 12 = −1.548

1.5

11 Determine (or approximate) the P-Value.

1.548

 -1.372 > -1.548 > -1.812

 .10 >

-value > .05

-value

18.4 One-Sample t Test

Example: Hemoglobin levels

Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

18.4 One-Sample t Test

Example: Hemoglobin levels

Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

18.4 One-Sample t Test

Example: Hemoglobin levels

Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is larger than 0.05, we fail to reject the null hypothesis.

18.4 One-Sample t Test

Example: Hemoglobin levels

Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim.

18.4 One-Sample t Test

Example: Hemoglobin levels

Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim.

There is not enough evidence (at a =.05) that, boys from this country are typically anemic (i.e., mean Hb <12 g/dl).

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (

unknown) TI-83/84:



STAT



TESTS

Select

Stats.

Enter m 0 s, 

TTest



Enter.

, 𝑥 n, and the confidence level.

Select

Calculate

 (Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.)

JMP:

Enter the data.

Analyze



Distribution.

“

Click-and-Drag

” (the appropriate variable)

into the

‘Y, Columns’

box.

Click

OK.

 

Click

on the red upside-down triangle next to the title of the variable from the

‘ Y,Columns ’

box.

Proceed to ‘

Confidence Interval

’ ->

Select the appropriate confidence level.

18.4 One-Sample t Test

Example: Hemoglobin levels (n= 11, 20, and 33)

▸ Let’s now use technology to conduct the test of significance at a = .05 for the three different sample sizes (n =11, 20, and 33). We will particularly focus on the test statistic, p-value, and decision at a = .05.

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance ( σ unknown)

▸

TI-83/84. STAT



TESTS



TTest



Enter.

  Select Stats. Enter 𝜇 0 , 𝑠, 𝑥 and n. Select Calculate .

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.) (for this example) 

Inpt >> STATS

  m

0 : 12 >> Calculate

: 11.3

(ENTER) >> s: 1.5 >>

n :

11 t= - 1.548 p= .076

m :

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance ( σ unknown)

▸

TI-83/84. STAT



TESTS



TTest



Enter.

  Select Stats. Enter 𝜇 0 , 𝑠, 𝑥 and n. Select Calculate .

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.) (for this example) 

Inpt >> STATS

  m

0 : 12 >>

: 11.3

Calculate (ENTER) >> s: 1.5 >>

n :

11 t= - 1.548 p= .076 >>

m :

< Fail to reject H 0, .076 > .05 => p-value >

There is not enough evidence (at a = .05) to conclude that boys from this country are typically anemic.

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance ( σ unknown)

▸

TI-83/84. STAT



TESTS



TTest



Enter.

  Select Stats. Enter 𝜇 0 , 𝑠, 𝑥 and n. Select Calculate .

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.) (for this example) 

Inpt >> STATS

  m

0 : 12 >> Calculate

: 11.3

(ENTER) >> s: 1.5 >>

n :

20 t= - 2.087 p= .025 >>

m :

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance ( σ unknown)

▸

TI-83/84. STAT



TESTS



TTest



Enter.

  Select Stats. Enter 𝜇 0 , 𝑠, 𝑥 and n. Select Calculate .

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.) (for this example) 

Inpt >> STATS

  m

0 : 12 >>

: 11.3

Calculate (ENTER) >> s: 1.5 >>

n :

20 t= - 2.087 p= .025 >>

m :

< Reject H 0, .025 < .05 => p-value <

There is enough evidence (at a = .05) to conclude boys from this country are typically anemic.

18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance ( σ unknown)

▸

TI-83/84. STAT



TESTS



TTest



Enter.

  Select Stats. Enter 𝜇 0 , 𝑠, 𝑥 and n. Select Calculate .

(Note: Select Data when 𝑥 and n are not provided. Then enter the list where the data are stored.) (for this example) 

Inpt >> STATS

  m

0 : 12 >>

: 11.3

Calculate (ENTER) >> s: 1.5 >>

n :

33 t= - 2.68 p= .006 >>

m :

< Reject H 0, .006 < .05 => p-value <

There is enough evidence (at a = .05) to conclude boys from this country are typically anemic.

18.6 Matched Pairs t Procedures

Matched Pairs t Procedures

▸ One way to demonstrate that a treatment causes an observed effect is to use a matched pairs experiment. ▸ In a

matched pairs design

 subjects are matched in pairs and each treatment is given to one subject in each pair or  observations are taken on the same subject

before-and-after

some treatment .

18.6 Matched Pairs t Procedures

Matched Pairs t Procedures

▸ To compare the responses to the two treatments in a matched pairs design, find the difference between the responses within each pair. Then apply the one-sample

procedures to these differences.

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Some researchers claim that music relaxes students and reduces stress while studying. 12 students were selected at random. Their initial resting pulse rate (beats/minute) was obtained, and each person participated in a month-long music-listening, relaxation therapy program. A final resting pulse rate was taken at the end of the experiment. The data are given below. ▸ Is there any evidence that music reduced the mean pulse rate, and consequently, reduced stress? Assume the underlying distributions are normal and use a 0.025 level of significance.

Subject Amy Bob Cal Dee Eve Fay Gus Hal Ike Joe Kel Moe

67 71 67 83 70 75 71 68 72 88 78 70

Initial Pulse Rate Final Pulse Rate

61 72 70 76 58 61 74 59 61 64 71 77

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:  D= “Diff” < 0 (implies that the pulse rate has “reduced.”)  Final Pulse Rate – Initial Pulse Rate < 0

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Compute the sample statistics for the Diff Final-Initial check the distribution for our assumptions.

Subject Initial Pulse Rate Final Pulse Rate D (final-initial) Amy Bob Cal

67 61

-6

71 72

67 70

3 Dee Eve Fay Gus Hal

83 70 75 71 68

Ike

72 76

-7

58 61

-12 -14

-9

Joe Kel

88 78

Moe

70 64

-11 -24

-7

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy Diff(Final-Initial)

Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N -6.3333 8.7316908 2.520622 -0.785482 -11.88118 12

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

State:

What is the practical question that requires a statistical test? Does music reduce stress (as measured by pulse rates)?

µ Final - Initial < 0 or µ D < 0 where D = Final Pulse Rate – Initial Pulse Rate

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Plan:

Identify the parameter.

m = mean difference between final and initial pulse rates

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Plan:

Identify the parameter.

m = mean difference between final and initial pulse rates b.

List all given information from the data collected . n=12, 𝑥 = −6.333

, s = 8.732

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Plan:

Identify the parameter.

m = mean difference between final and initial pulse rates b.

List all given information from the data collected . n=12, 𝑥 = −6.333

, s = 8.732

State the null (H 0 ) and alternative (H a ) hypotheses.

H 0 : m Diff = 0 Specify the level of significance . a = .025

H A : m Diff < 0

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Plan:

Identify the parameter.

m = mean difference between final and initial pulse rates b.

List all given information from the data collected . n=12, 𝑥 = −6.333

, s = 8.732

State the null (H 0 ) and alternative (H a ) hypotheses.

H 0 : m Diff = 0 Specify the level of significance . a = .025

H A : m Diff < 0 e.

Determine the type of test.

Left-tailed Right-tailed Sketch the region(s) of “extremely unlikely” test statistics.

Two-Tailed

Example: Music relaxation therapy Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

18.4 One-Sample t Test

Example: Music relaxation therapy Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

 Large enough population: sample ratio?

Yes.

The number of students is arbitrarily large; therefore, N > 20*12 = 240.

18.4 One-Sample t Test

Example: Music relaxation therapy Solve:

a) Check the conditions for the test you plan to use.  Random Sample?

Yes.

Stated as a random sample.

 Large enough population: sample ratio?

Yes.

The number of students is arbitrarily large; therefore, N > 20*12 = 240.

 Large enough sample; Normal or t-distribution?

Yes.

n = 12 < 40. But the data seem to be approximately Normal, so we will attempt to use the t-distribution.

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝜎 𝑛 −6.33 − 0 = 8.732

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝜎 𝑛 = −6.33 − 0 8.732

12 = −2.511

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) c) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝜎 𝑛 = −6.33 − 0 8.732

12 = −2.511

Determine (or approximate) the P-Value.

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝜎 𝑛 = −6.33 − 0 8.732

12 = −2.511

c) Determine (or approximate) the P-Value.

 -2.328 > -2.511 >-2.718

2.511

18.4 One-Sample t Test

Example: Hemoglobin levels Solve:

b) Calculate the test statistic 𝑡 = 𝑥 − 𝜇 0 𝜎 𝑛 = −6.33 − 0 8.732

12 = −2.511

c) Determine (or approximate) the P-Value.

 -2.328 > -2.511 >-2.718



.02 > P-Value >.01

2.511

-value

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Conclude:

a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Conclude:

a) Make a decision about the null hypothesis ( Reject H 0 or Fail to reject H 0 ).

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Conclude:

a) Make a decision about the null hypothesis ( Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is less than 0.025, we reject the null hypothesis.

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Conclude:

a) Make a decision about the null hypothesis ( Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim.

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸

Conclude:

a) Make a decision about the null hypothesis ( Reject H 0 or Fail to reject H 0 ). Because the approximate

-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim.

There is enough evidence at the a =.025 level to conclude that music reduces stress (and lowers pulse rates).

18.6 Matched Pairs t Procedures

Question

▸ Based upon our data, are matched pairs based upon dependent samples?

a) b) Yes. (The data from 2 nd variable are related to the data from the 1 st variable).

No. (The data from 2 nd variable are not related to the data from the 1 st variable).

c) Not sure

18.6 Matched Pairs t Procedures

Question

▸ Based upon our data, are matched pairs based upon dependent samples?

a) b) Yes. (The data from 2 nd variable are related to the data from the 1 st variable).

No. (The data from 2 nd variable are not related to the data from the 1 st variable).

c) Not sure

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find

t*(11, .95)=2.201

by looking across the df = 11 row and down the 95% confidence level column.

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find

t*(11, .95)=2.201

by looking across the df = 11 row and down the 95% confidence level column.





s n

=   8.732

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find

t*(11, .95)=2.201

by looking across the df = 11 row and down the 95% confidence level column.





s n

=   8.732

=   12 =  0.785 to -11.881

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find

t*(11, .95)=2.201

by looking across the df = 11 row and down the 95% confidence level column.





s n

=   8.732

=   12 =  0.785 to -11.881

We are 95% confident that the mean difference in final and initial pulse rates (with Final minus Initial) is between -0.785 and -11.881 bpm. Or, we are 95% confident that participating in the relaxation experiment reduce one’s pulse rate by an average of 0.7852 to 11.881 bpm.

18.6 Matched Pairs t Procedures

Constructing a 95% confidence interval for the average difference in initial and final pulse rates.

▸ We’ll use the TI-83/84 first this time. Then verify via the formula.

▸ Enter the differences { -6, 1, 3, -7, -12, -14, 3, -9, -11, -24, -7, 7} in L 1 .



STAT >> TESTS >> TInterval >> Enter

 Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.



Inpt >> DATA



List: L 1



C-Level : >> 95

Freq: 1 

Calculate (ENTER)

18.6 Matched Pairs t Procedures

Constructing a 95% confidence interval for the average difference in initial and final pulse rates.

▸ TI-83/84 ▸ Enter the differences { -6, 1, 3, -7, -12, -14, 3, -9, -11, -24, -7, 7} in L 1 .



STAT >> TESTS >> TInterval >> Enter

 Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.



Inpt >> DATA



List: L 1



C-Level : >> 95

Freq: 1 

Calculate (ENTER) ( -11.88, -0.7855)

18.6 Matched Pairs t Procedures

Constructing a 95% confidence interval for the average difference in initial and final pulse rates.

▸ TI-83/84  

STAT >> TESTS >> TInterval >> Enter

Note: Select

Data

when 𝑥 and 𝑛 are not provided. Then enter the list where the data are stored.

(for this example) 

Inpt >> STATS

 : -6.33

>> s: automatically) 8.732



C-Level : 95



Calculate (ENTER)

: -6.33

n :

12 ( -11.88, -0.7855) (these should populate

18.6 Matched Pairs t Procedures

Example: Music relaxation therapy

▸ Technology output for the hypothesis test and confidence interval:

Technology output for the hypothesis test and confidence interval: Test Mean

Hypothesized Value Actual Estimate DF Std Dev Test Statistic Prob > |t| Prob > t Prob < t

t Test

-2.5126 0.0289* 0.9856 0.0144* 0 -6.3333 11 8.73169

Confidence Intervals Parameter Estimate

Mean Std Dev -6.33333 8.731691

Lower CI

-11.8812 6.185487

Upper CI 1-Alpha

-0.78548 0.950 14.82535 0.950

18.7 Robustness of t Procedures

Robustness of t Procedures

▸ A confidence interval or significance test is called

robust

if the confidence interval or P-value does not change very much when the conditions for use of the procedure are violated.

▸ Except in the case of small samples, the condition that the data are an SRS from the population of interest is more important than the condition that the population distribution is Normal.

18.7 Robustness of t Procedures

Robustness of t Procedures

▸ The t-procedures guard against non-Normality except when there is strong skewness or outliers present. ▸ When the data are not from a Normal distribution we also need to consider the sample size: Sample size less than 15 Sample size between 15 and 40 Sample size is at least 40 The

procedures can be used if the data close to Normal (roughly symmetric, single peak, no outliers)? If there is

clear skewness

outliers

then,

do not use t

The

procedures can be used

except in the presences of outliers

or strong skewness.

The

t procedures can be used

even for clearly skewed distributions.

18.7 Robustness of t Procedures

Robustness of t Procedures

▸ Note that we have changed the “large enough sample” condition to be adaptable to the situations that we encounter. This is because

procedures are

robust

against violations of Normality.

18.7 Robustness of t Procedures

Robustness of t Procedures

▸ Example: The number of text messages sent daily for 25 college students are below. Can we safely use

-procedures?

10 10 12 15 17 18 22 23 24 25 25 25 26 27 28 30 103 42 118 51 120 53 130 75 135

18.7 Robustness of t Procedures

Robustness of t Procedures

▸ Example: The number of text messages sent daily for 25 college students are below. Can we safely use

-procedures?

10 10 12 15 17 18 22 23 24 25 25 25 26 27 28 30 103 42 118 51 120 53 130 75 135 ▸ No. The sample size is 25, but there are clear outliers and strong skewness.

Five-Minute Summary

▸ List at least 3 concepts that had the most impact on your knowledge of inference about a population mean.

_____________ _________________ _______________

Chapter 3: The Normal Distributions

Transcript Chapter 3: The Normal Distributions

Chapter 18: Inference about One Population Mean

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