Download 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
Sample Statistic
and Point Estimate
Population
Parameter
Mean
𝑥
μ
s
σ
𝑝
𝑝
Standard
Deviation
Proportion
of
Successes
▸ 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 σ will be known and
the population mean μ will not be known.
▸ Chapter 14 taught us that 𝑥 is the best point estimate of µ.
Similarly, s can estimate σ.
18.0 Inference about One Population Mean
Inference when σ is unknown
▸ In Chapter 11 when σ was known we used
𝑥−𝜇
𝑧=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 t-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 n from a large population that has the Normal
distribution with mean μ and standard deviation σ.
The one-sample t statistic
𝑥−𝜇
𝑡=𝑠
𝑛
has the t distribution with n – 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 t 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 t distributions (dashed) and the standard Normal
distribution (solid):
18.2 The t Distributions
T-distribution Compared to the Z-Distribution
Similarities
Differences
Symmetric about 0
T has thicker tails
Single-peaked & Bellshaped
Varies based upon
‘degrees of freedom’
18.2 The t Distributions
T-distribution Compared to the Z-Distribution
▸ Poll: The t2 curve has thicker dashes. The t9 curve has smaller
dashes.
Z is the solid curve. What would you anticipate happening to
tdf as the degrees of freedom (df) increase?
a) tdf will not be affected
b) tdf will approach Z
c) tdf will become further from Z.
18.2 The t Distributions
T-distribution Compared to the Z-Distribution
▸ Poll: The t2 curve has thicker dashes. The t9 curve has smaller
dashes.
Z is the solid curve. What would you anticipate happening to
tdf as the degrees of freedom (df) increase?
a) tdf will not be affected
b) tdf will approach Z
c) tdf 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 n from a large population having unknown mean
μ.
▸ A level C confidence interval for μ is
𝑥±
𝑡∗
𝑠
𝑛
▸ where t* is the critical value for the t(n – 1) density curve with area C
between ‒ t* and t*. This interval is exact when the population
distribution is Normal and is approximately correct for large n 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
𝑥 ±(1.96 or another z-score)
𝑥
± 𝑡
∗
𝑠
𝑛
Introduction of Confidence Intervals
𝜎
𝑛
General Form (when σ is known)
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?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
1. Components
Do we have an SRS?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
Large enough population: sample ratio?
Yes.
.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
1. Components
Do we have an SRS?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
Large enough population: sample ratio?
Yes.
N > 20*n=20*11=220
N=Population of boys > 220.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
1. Components
Do we have an SRS?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
Large enough population: sample ratio?
Yes.
N > 20*n=20*11=220
N=Population of boys > 220.
t-distribution?
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
1. Components
Do we have an SRS?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
Large enough population: sample ratio?
Yes.
N > 20*n=20*11=220
N=Population of boys > 220.
t-distribution?
Yes.
n = 11 < 40 …
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
1. Components
Do we have an SRS?
Yes. Stated as a random sample.
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
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.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
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.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
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.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3.
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 t*.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
t*(90%, 10) = 1.812
3.
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 t*.
90%
df=11-1=10
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
t*(90%, 10) = 1.812
3.
4.
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 t*.
Construct the confidence interval.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
t*(90%, 10) = 1.812
4. Interval.
11.3 ± 1.812
1.5
11
=
3.
4.
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 t*.
Construct the confidence interval.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
t*(90%, 10) = 1.812
4. Interval.
11.3 ± 1.812
1.5
11
= 11.3 ± .82
3.
4.
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 t*.
Construct the confidence interval.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
4. Interval.
11.3 ± 1.812
t*(90%, 10) = 1.812
1.5
11
3.
4.
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 t*.
Construct the confidence interval.
= 11.3 ± .82 = (10.48,12.12)
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
4. Interval.
11.3 ± 1.812
t*(90%, 10) = 1.812
1.5
11
3.
4.
5.
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 t*.
Construct the confidence interval.
*Interpret* the interval.
= 11.3 ± .82 = (10.48,12.12)
18.3 One-Sample t Confidence Intervals
Example: 90% CI for Hb levels
2. Components.
𝒙 = 11.3, s = 1.5, n = 11
Steps for SuccessConstructing Confidence Intervals for m
(s unknown).
1.
2.
3. Select t*.
df =(n-1)=10
4. Interval.
11.3 ± 1.812
t*(90%, 10) = 1.812
1.5
11
3.
4.
5.
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 t*.
Construct the confidence interval.
*Interpret* the interval.
= 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 (s 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 on 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
 𝒙 : 11.3 >> s: 1.5

C-Level : 90
 Calculate (ENTER)
>> n : 11
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
 𝒙 : 11.3 >> s: 1.5

>> n : 11
C-Level : 90
 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
 𝒙 : 11.3 >> s: 1.5

>> n : 20
C-Level : 90
 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
 𝒙 : 11.3 >> s: 1.5

C-Level : 90
 Calculate (ENTER)
>> n : 33
Larger n, narrower interval.
(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
𝑥−𝜇
𝑡=𝑠
𝑛
has the t distribution with n – 1 degrees of freedom.
18.4 One-Sample t Test
The One-sample t Test of Significance
▸ To test the hypothesis
𝐻𝑜 : 𝜇 = 𝜇𝑜 , compute the one-sample t statistic
𝑥 − 𝜇𝑜
𝑡= 𝑠
𝑛
18.4 One-Sample t Test
The One-sample t Test of Significance
▸ The p-value for a test of H0 against
 𝐻𝑎 : 𝜇 < 𝜇0 is 𝑃 𝑇 ≤ 𝑡 .
 𝐻𝑎 : 𝜇 > 𝜇0 is 𝑃(𝑇 ≥ 𝑡).
 𝐻𝑎 : 𝜇 = 𝜇0 is 2𝑃(𝑇 ≥ 𝑡 ).
 These P-values are exact if the population distribution is Normal and are
approximately correct for large n 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
▸ 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 ___anemia________?
15.4 Tests for a Population Mean
Steps for Success –
Conducting Tests of Significance
Steps for SuccessConducting 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 (H0) and alternative (HA) 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 (H0) and alternative (HA) hypotheses.
H0: µ = 12
Ha :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 (H0) and alternative (HA) hypotheses.
d) Specify the level of significance. α =.05
H0: µ = 12
Ha :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 (H0) and alternative (HA) hypotheses.
H0: µ = 12
Ha :µ < 12
d) Specify the level of significance. α =.05
e) Determine the type of test.
Left-tailed
Right-tailed
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 (H0) and alternative (HA) hypotheses.
H0: µ = 12
Ha :µ < 12
d) Specify the level of significance. α =.05
e) Determine the type of test.
Left-tailed
Right-tailed
Two-Tailed
18.4 One-Sample t Test
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.
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
b)
Calculate the test statistic
𝑥 − 𝜇0
𝑡=
=
𝑠 𝑛
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
b)
Calculate the test statistic
𝑥 − 𝜇0 11.3 − 12
𝑡=
=
= −1.548
𝑠 𝑛
1.5 11
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
b)
Calculate the test statistic
c)
Determine (or approximate) the P-Value. 1.548
𝑥 − 𝜇0 11.3 − 12
𝑡=
=
= −1.548
𝑠 𝑛
1.5 11
 -1.372 > -1.548 > -1.812
P-value
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
b)
Calculate the test statistic
c)
Determine (or approximate) the P-Value. 1.548
𝑥 − 𝜇0 11.3 − 12
𝑡=
=
= −1.548
𝑠 𝑛
1.5 11
 -1.372 > -1.548 > -1.812
 .10 > P-value > .05
P-value
18.4 One-Sample t Test
Example: Hemoglobin levels
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
18.4 One-Sample t Test
Example: Hemoglobin levels
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
18.4 One-Sample t Test
Example: Hemoglobin levels
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
Because the approximate P-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 H0 or Fail to reject H0).
Because the approximate P-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 H0 or Fail to reject H0).
Because the approximate P-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 (s unknown)
TI-83/84: STAT  TESTS  TTest  Enter.
 Select Stats. Enter m0 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 on 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 𝜇0 , 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
 m0: 12 >> 𝒙 : 11.3 >> s: 1.5
>> n : 11
>> m : <
 Calculate (ENTER)
t= - 1.548 p= .076
18.4 One-Sample t Test
Technology Tips –
Conducting Tests of Significance (σ unknown)
▸ TI-83/84. STAT  TESTS  TTest  Enter.
 Select Stats. Enter 𝜇0 , 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
 m0: 12 >> 𝒙 : 11.3 >> s: 1.5
 Calculate (ENTER)
>> n : 11
>> m : <
Fail to reject H0, .076 > .05
=> p-value > a .
There is not enough evidence (at a = .05) to conclude
that boys from this country are typically anemic.
t= - 1.548 p= .076
18.4 One-Sample t Test
Technology Tips –
Conducting Tests of Significance (σ unknown)
▸ TI-83/84. STAT  TESTS  TTest  Enter.
 Select Stats. Enter 𝜇0 , 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
 m0: 12 >> 𝒙 : 11.3 >> s: 1.5
>> n : 20
 Calculate (ENTER)
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 𝜇0 , 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
 m0: 12 >> 𝒙 : 11.3 >> s: 1.5
 Calculate (ENTER)
>> n : 20
>> m : <
Reject H0, .025 < .05
=> p-value < a .
There is enough evidence (at a = .05) to conclude
boys from this country are typically anemic.
t= - 2.087 p= .025
18.4 One-Sample t Test
Technology Tips –
Conducting Tests of Significance (σ unknown)
▸ TI-83/84. STAT  TESTS  TTest  Enter.
 Select Stats. Enter 𝜇0 , 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
 m0: 12 >> 𝒙 : 11.3 >> s: 1.5
 Calculate (ENTER)
>> n : 33
>> m : <
Reject H0, .006 < .05
=> p-value < a .
There is enough evidence (at a = .05) to conclude
boys from this country are typically anemic.
t= - 2.68 p= .006
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 t 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
Initial Pulse
Rate
Final Pulse
Rate
Amy Bob Cal
Dee Eve Fay Gus Hal
Ike
Joe Kel
Moe
67
71
67
83
70
75
71
68
72
88
78
70
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
▸ 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.”)
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
▸ 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
 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.:
 D= “Diff” < 0 (implies that the pulse rate has “reduced.”)
 Final Pulse Rate – Initial Pulse Rate < 0
 Final Pulse Rate < Initial Pulse Rate
 DiffFinal-Initial= DiffFinal<Initial this is preferred since it directly aligns with the
hypothesis presented.
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Compute the sample statistics for the DiffFinal-Initial check the distribution
for our assumptions.
Subject
Amy Bob Cal
Dee Eve Fay Gus Hal
Ike
Joe Kel
Moe
Initial Pulse
Rate
Final Pulse
Rate
67
71
67
83
70
75
71
68
72
88
78
70
61
72
70
76
58
61
74
59
61
64
71
77
1
3
D (final-initial)
-6
-7 -12 -14
3
-9 -11 -24 -7
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)?
µFinal - Initial < 0
or
µD < 0 where D = Final Pulse Rate – Initial Pulse Rate
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Plan:
a. Identify the parameter. m = mean difference between final and initial pulse rates
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Plan:
a. 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:
a. 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
c. State the null (H0) and alternative (Ha) hypotheses. H0: mDiff = 0
d. Specify the level of significance.
a = .025
HA: mDiff < 0
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Plan:
a. 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
c. State the null (H0) and alternative (Ha) hypotheses. H0: mDiff = 0
d. Specify the level of significance.
HA: mDiff < 0
a = .025
e.
Determine the type of test.
Left-tailed
Right-tailed
f.
Sketch the region(s) of “extremely unlikely” test statistics.
Two-Tailed
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.
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
𝑥 − 𝜇0
−6.33 − 0
𝑡=
=
𝜎 𝑛 8.732 12
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
b)
Calculate the test statistic
𝑥 − 𝜇0
−6.33 − 0
𝑡=
=
= −2.511
𝜎 𝑛 8.732 12
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
𝑥 − 𝜇0
−6.33 − 0
𝑡=
=
= −2.511
𝜎 𝑛 8.732 12
b)
Calculate the test statistic
c)
Determine (or approximate) the P-Value.
18.4 One-Sample t Test
Example: Hemoglobin levels
Solve:
𝑥 − 𝜇0
−6.33 − 0
𝑡=
=
= −2.511
𝜎 𝑛 8.732 12
b)
Calculate the test statistic
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:
𝑥 − 𝜇0
−6.33 − 0
𝑡=
=
= −2.511
𝜎 𝑛 8.732 12
b)
Calculate the test statistic
c)
Determine (or approximate) the P-Value.
2.511
 -2.328 > -2.511 >-2.718

.02 > P-Value >.01
P-value
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
18.6 Matched Pairs t Procedures
Example: Music relaxation therapy
▸ Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
Because the approximate P-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 H0 or Fail to reject H0).
Because the approximate P-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 H0 or Fail to reject H0).
Because the approximate P-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)
Yes. (The data from 2nd variable are related to the data from the 1st
variable).
b)
No. (The data from 2nd variable are not related to the data from the 1st
variable).
c)
Not sure
18.6 Matched Pairs t Procedures
Question
▸ Based upon our data, are matched pairs based upon dependent
samples?
a)
Yes. (The data from 2nd variable are related to the data from the 1st
variable).
b)
No. (The data from 2nd variable are not related to the data from the 1st
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.
x  t
s
8.732
= 6.333  2.201
n
12
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.
x  t
s
8.732
= 6.333  2.201
= 6.333  5.548 = 0.785 to -11.881
n
12
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.
x  t
s
8.732
= 6.333  2.201
= 6.333  5.548 = 0.785 to -11.881
n
12
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 L1.
 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: L1
>> Freq: 1
 C-Level : 95
 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 L1.
 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: L1
>> Freq: 1
 C-Level : 95
 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
 𝒙 : -6.33
>> s: 8.732 >> 𝒙 : -6.33 >> n : 12
automatically)

C-Level : 95
 Calculate (ENTER)
( -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
0
Actual Estimate
-6.3333
DF
11
Std Dev
8.73169
t Test
Test Statistic
-2.5126
Prob > |t|
0.0289*
Prob > t
0.9856
Prob < t
0.0144*
Confidence Intervals
Parameter
Estimate
Mean
-6.33333
Std Dev
8.731691
Lower CI
-11.8812
6.185487
Upper CI
-0.78548
14.82535
1-Alpha
0.950
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
The t procedures can be used if the data close to Normal
(roughly symmetric, single peak, no outliers)? If there is
clear skewness or outliers then, do not use t.
Sample size between 15
and 40
The t procedures can be used except in the presences
of outliers or strong skewness.
Sample size is at least 40
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 t
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 t-procedures?
10
10
12
15
17
18
22
23
24
25
25
25
26
27
28
30
42
51
53
75
103
118
120
130
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 t-procedures?
10
10
12
15
17
18
22
23
24
25
25
25
26
27
28
30
42
51
53
75
103
118
120
130
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.
_____________
_________________
_______________