COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 9: The t Statistic.

COURSE: JUST 3900TIPS FOR APLIA

Developed By: Ethan Cooper (Lead Tutor)

John LohmanMichael Mattocks

Aubrey Urwick

Chapter 9: The t Statistic

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: t-Score Formula: estimated Cohen’s d: r2:

You May Need These Formulas

Sample Mean: Definitional Formula for SS: Computational Formula for SS: Sample Variance: Sample Standard Deviation:

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

Question 1 Answer: 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) Compute the variance for the sample.

b) Compute the estimated standard error for the sample mean.

Estimated Standard Error

Question 2 Answer:

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

Question 3 Answer:

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)


Question 4 Answer: 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


t = - 3.182 t = + 3.182

df = 3

t Distribution with α = 0.05 Critical regionCritical region


Question 4 Answer: Step 3: Compute sample statistic

a)


Critical regionCritical region

t = - 3.182 t = + 3.182

df = 3

t Distribution with α = 0.05

t = 2.00


Question 4 Answer Step 4: Make a decision

For a Two-tailed Test:

tsample (2.00) < tcritical (3.182) Thus, we fail to reject the null and cannot conclude that the

treatment has an effect.

If -3.182 < tsample < 3.182, fail to reject H0

If tsample ≤ -3.182 or tsample ≥ 3.182, reject H0



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


t Distribution with α = 0.05

df = 15

t = - 2.131 t = + 2.131

Critical region Critical region



a)


t Distribution with α = 0.05Critical region

t = - 2.131 t = + 2.131

Critical region

df = 15

t = 4.00


Question 4 Answer: Step 4: Make a decision

For a Two-tailed Test:

tsample (4.00) < tcritical (2.131) Thus, we reject the null and conclude that the treatment has an

effect.

If -2.131 < tsample < 2.131, fail to reject H0

If tsample ≤ -2.131 or tsample ≥ 2.131, reject H0

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.



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


df = 24


t = -1.711

Because this is a one-tailed test‚there is only one critical region.



a)


df = 24


t = -1.711

Because this is a one-tailed test‚there is only one critical region.

t = -1.00

One-tailed Hypothesis Test

Question 5 Answer: Step 4: Make a decision

For a One-tailed Test:

tsample (-1.00) > tcritical (-1.711) Thus, we fail to reject the null and cannot conclude that the

supplement decreases 40 yard dash times.

If tsample > -1.711, fail to reject H0

If tsample ≤ -1.711, reject H0

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) Find estimated Cohen’s d.

b) Find r2.


Question 6a Answer: estimated Cohen’s d: d = This is a medium to large effect.

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.


Question 6b Answer:

Find t and df.

𝑑𝑓=𝑛−1=16−1=15

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.


Question 7 Answer: 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


Middle 95%of t distribution

t = + 2.131t = - 2.131

df = 15

2.5% in the lower tail 2.5% in the upper tail


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*s

M

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


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.

M = 8681.738 90.262

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

Frequently Asked Questions 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.


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.


Sample Mean: Definitional Formula for SS:

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


Sample Variance: Sample Standard Deviation: Estimated Standard Error:

COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 9: The t Statistic.

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