• Study Resource
• Explore

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Student's t-test wikipedia, lookup

Taylor's law wikipedia, lookup

Bootstrapping (statistics) wikipedia, lookup

Resampling (statistics) wikipedia, lookup

Misuse of statistics wikipedia, lookup

Psychometrics wikipedia, lookup

Foundations of statistics wikipedia, lookup

Omnibus test wikipedia, lookup

Sufficient statistic wikipedia, lookup

Transcript
```COURSE: JUST 3900
TIPS FOR APLIA
Chapter 9:
The t Statistic
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Key Terms: Don’t Forget
Notecards











Hypothesis Test (p. 233)
Null Hypothesis (p. 236)
Alternative Hypothesis (p. 236)
Alpha Level (level of significance) (pp. 238 & 245)
Critical Region (p. 238)
Estimated Standard Error (p. 286)
t statistic (p. 286)
Degrees of Freedom (p. 287)
t distribution (p. 287)
Confidence Interval (p. 300)
Directional (one-tailed) Hypothesis Test (p. 304)
Formulas

Estimated Standard Error: 𝑠𝑀 =
𝑠
𝑛
=

t-Score Formula: 𝑡 =
𝑀−𝜇
𝑠𝑀

estimated Cohen’s d:
𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝑠
 r 2:

𝑟2
=
𝑡2
𝑡 2 +𝑑𝑓
𝑑𝑓 = 𝑛 − 1
𝑠2
𝑛
=
=
𝑠2
𝑛
You May Need These Formulas
𝑋
𝑛

Sample Mean: 𝑀 =

Definitional Formula for SS: 𝑆𝑆 =



𝑋−𝑀
𝑋2
Computational Formula for SS: 𝑆𝑆 =
Sample Variance:
𝑠2
=
2
−
𝑆𝑆
𝑛−1
Sample Standard Deviation: 𝑠 =
𝑠2
=
𝑆𝑆
𝑛−1
𝑋 2
𝑛
When to Use the t Statistic

Question 1: Under what circumstances is a t statistic
used instead of a z-score for a hypothesis test?
When to Use the t Statistic


A t statistic is used instead of a z-score when the population
standard deviation (σ) and variance (σ2) are not known.
Estimated Standard Error

Question 2: A sample of n = 9 scores has a SS = 288.
a)
b)
Compute the variance for the sample.
Compute the estimated standard error for the sample mean.
Estimated Standard Error

a)
b)
𝑠2 =
𝑆𝑆
𝑛−1
=
𝑠𝑀 =
𝑠2
𝑛
=
288
9−1
36
9
=
288
8
= 36
= 4=2
Possible Outcomes of a
Hypothesis Test

Question 3: A researcher reports a t statistic with df = 20.
How many individuals participated in the study?
Possible Outcomes of a
Hypothesis Test




𝑑𝑓 = 𝑛 − 1
20 = 𝑛 − 1
𝑛 = 21
Two-Tailed Hypothesis Test
Using the t Statistic

Question 4: A sample of n = 4 is selected from a
population with a mean of µ = 40. A treatment is
administered to the individuals in the sample and, after
treatment, the sample has a mean of M = 44 and a
variance of s2 = 16.


Is this sample sufficient to conclude that the treatment has an
effect? (Use a two-tailed test with α = 0.05)
If all other factors are held constant and the sample size is
increased to n = 16, is the sample sufficient to conclude that the
treatment has a significant effect? (Use a two-tailed test with
α = 0.05)
Two-Tailed Hypothesis Test
Using the t Statistic


Step 1: State hypotheses



H0: Treatment has no effect. (µ = 40)
H1: Treatment has an effect. (µ ≠ 40)
Step 2: Set Criteria for Decision (α = 0.05)
Critical t = ± 3.182
Two-Tailed Hypothesis Test
Using the t Statistic
df = 3
t Distribution with
α = 0.05
Critical region
t = - 3.182
Critical region
t = + 3.182
Two-Tailed Hypothesis Test
Using the t Statistic


Step 3: Compute sample statistic
𝑠2
𝑛
=
𝑀−𝜇
𝑠𝑀
=
a)
𝑠𝑀 =
b)
𝑡=
16
4
=
44−40
2
4=2
=
4
2
= 2.00
Two-Tailed Hypothesis Test
Using the t Statistic
df = 3
t Distribution with
α = 0.05
Critical region
t = - 3.182
t = 2.00
Critical region
t = + 3.182
Two-Tailed Hypothesis Test
Using the t Statistic


Step 4: Make a decision

For a Two-tailed Test:
If -3.182 < tsample < 3.182, fail to reject H0
If tsample ≤ -3.182 or tsample ≥ 3.182, reject H0


tsample (2.00) < tcritical (3.182)
Thus, we fail to reject the null and cannot conclude that the
treatment has an effect.
Two-Tailed Hypothesis Test
Using the t Statistic


Step 1: State hypotheses



H0: Treatment has no effect. (µ = 40)
H1: Treatment has an effect. (µ ≠ 40)
Step 2: Set Criteria for Decision (α = 0.05)
Critical t = ± 2.131
Two-Tailed Hypothesis Test
Using the t Statistic
df = 15
t Distribution with
α = 0.05
Critical region
t = - 2.131
Critical region
t = + 2.131
Two-Tailed Hypothesis Test
Using the t Statistic


Step 3: Compute sample statistic
a)
𝑠𝑀 =
b)
𝑡=
16
16
𝑀−𝜇
𝑠𝑀
=
=
1=1
44−40
1
=
4
1
= 4.00
Two-Tailed Hypothesis Test
Using the t Statistic
df = 15
t Distribution with
α = 0.05
Critical region
t = - 2.131
Critical region
t = + 2.131
t = 4.00
Two-Tailed Hypothesis Test
Using the t Statistic


Step 4: Make a decision

For a Two-tailed Test:
If -2.131 < tsample < 2.131, fail to reject H0
If tsample ≤ -2.131 or tsample ≥ 2.131, reject H0


tsample (4.00) < tcritical (2.131)
Thus, we reject the null and conclude that the treatment has an
effect.
One-Tailed Hypothesis Test
Using the t Statistic

Question 5: Dr. Johnson is conducting research on a
new supplement that claims to improve physical
performance. The average 40 yard dash time for college
football players is µ = 4.65 seconds. A sample of n = 25
players is chosen to test this supplement. Each player is
given the supplement, and, 1 hour later, 40 yard dash
time is measured for each player. The average time for
this sample is M = 4.57 seconds with SS = 3.6. Dr.
Anderson would like to use a hypothesis test with
α = 0.05 to evaluate the effectiveness of the supplement.

Use a one-tailed test to determine whether the supplement
produces a significant decrease in 40 yard dash times.
One-Tailed Hypothesis Test
Using the t Statistic


Step 1: State hypotheses



H0: Supplement does not decrease 40 yard dash times. (µ ≥ 4.65)
H1: Supplement does decrease 40 yard dash times. (µ < 4.65)
Step 2: Set Criteria for Decision (α = 0.05)
Critical t: -1.711
One-Tailed Hypothesis Test
Using the t Statistic
df = 24
t Distribution with
α = 0.05
Critical region
Because this is a
one-tailed test‚
there is only one
critical region.
t = -1.711
One-Tailed Hypothesis Test
Using the t Statistic


Step 3: Compute sample statistic
a)
𝑠2 =
b)
𝑠𝑀 =
c)
𝑡=
𝑆𝑆
𝑛−1
=
0.15
25
𝑀−𝜇
𝑠𝑀
3.6
25−1
=
=
=
3.6
24
= 0.15
0.006 = 0.08
4.57−4.65
0.08
=
−0.08
0.08
= −1.00
One-Tailed Hypothesis Test
Using the t Statistic
df = 24
t Distribution with
α = 0.05
Critical region
Because this is a
one-tailed test‚
there is only one
critical region.
t = -1.711
t = -1.00
One-tailed Hypothesis Test


Step 4: Make a decision

For a One-tailed Test:
If tsample > -1.711, fail to reject H0
If tsample ≤ -1.711, reject H0


tsample (-1.00) > tcritical (-1.711)
Thus, we fail to reject the null and cannot conclude that the
supplement decreases 40 yard dash times.
Estimated Cohen’s d and r2

Question 6: A sample of n = 16 is selected from a
population with a mean of µ = 80. A treatment is
administered to the sample and, after treatment, the
sample is found to be M = 86 with a standard deviation
of s = 8.
a)
b)
Find estimated Cohen’s d.
Find r2.
Estimated Cohen’s d and r2

𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

estimated Cohen’s d:

d=

This is a medium to large effect.
86−80
8
=
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝑠
6
8
= = 0.75
Remember: This week in Aplia, if d falls between 0.2 and 0.5, it is a small to medium
effect. If d falls between 0.5 and 0.8, it is a medium to large effect.
Estimated Cohen’s d and r2

 𝑟2
=
𝑡2
𝑡 2 +𝑑𝑓
𝑑𝑓 = 𝑛 − 1 = 16 − 1 = 15

Find t and df.

𝑠𝑀 =

𝑡=

𝑟 2 = 𝑡 2 +𝑑𝑓 = 32+15 = 9+15 = 24 = 0.375 = 37.5%
𝑠
𝑛
=
𝑀−𝜇
𝑠𝑀
=
𝑡2
8
16
8
= 4 = 2.00
86−80
2
32
=
6
2
= 3.00
9
9
This is a large effect.
Confidence Intervals

Question 7: A sample of n = 16 is selected from a
population with a mean of µ = 80. A treatment is
administered to the sample and, after treatment, the
sample is found to be M = 86 with a standard deviation
of s = 8.

Find the 95% confidence interval for the population mean after
treatment.
Confidence Intervals





Step #1: Look up the corresponding t values in the t distribution
table for scores that crop the middle 95% of the distribution.
This means that you need to have 2.5% in each tail.
Calculate the degrees of freedom for t : df = n – 1 = 16 -1 = 15
Now look up the values of t with 15 df for a 1 tail test at 0.025 or
a 2 tail test at 0.05: t = +/- 2.131
Confidence Intervals
df = 15
Middle 95%
of t distribution
2.5% in the lower tail
t = - 2.131
2.5% in the upper tail
t = + 2.131
Confidence Intervals

Step #2: Calculate the bounded values of the interval.
To do this, you must use M and sM as obtained from the
sample data and plug these values into the estimation
formula: μ = M ± t*sM
μ = M ± t*sM = 86 +/-
(2.131)*(2.00)
μlower = M - t*sM = 86 - (2.131)*(2.00) = 81.738
μupper = M + t*sM = 86 + (2.131)*(2.00) = 90.262
Confidence Intervals

Step #3: Summarize the findings
After treatment our population mean should fall between
μ = 81.738 and μ = 90.262. We are 95% confident that
the true population mean is located within this interval.
Value for lower
boundary of 95%
confidence interval
Value for upper
boundary of 95%
confidence interval
Values that fall in the
middle of the 95% CI
81.738
M = 86
90.262
FAQs

Will we be asked to remember formulas and concepts
from past chapters?


Like all math classes, statistics is a course that builds on itself as
the semester progresses. Therefore, it is entirely possible that
either Aplia, the test, or both will expect you to remember
formulas and concepts from previous chapters.
There is an example on the next slide.
FAQs

The following sample was obtained from a population
with unknown parameters.

Scores: 6, 12, 0, 3, 4


Compute the sample mean and standard deviation.
Compute the estimated standard error for M.
FAQs
𝑋
𝑛
=
6+12+0+3+4
5

Sample Mean: 𝑀 =

Definitional Formula for SS: 𝑆𝑆 =
=
25
5
=5
𝑋−𝑀
2
X
X-M
(X – M)2
6
6–5=1
(1)2 = 1
12
12 – 5 = 7
(7)2 = 49
0
0 – 5 = -5
(-5)2 = 25
3
3 – 5 = -2
(-2)2 = 4
4
4 – 5 = -1
(-1)2 = 1
SS = 80
FAQs
𝑆𝑆
𝑛−1
80
5−1

Sample Variance: 𝑠 2 =

Sample Standard Deviation: 𝑠 =

=
Estimated Standard Error: 𝑠𝑀 =
=
80
4
= 20
𝑠2 =
𝑠2
𝑛
=
20
5
20 = 4.47
= 2.00
```
Related documents