1. WARM – UP (1)
What would be the critical value of t for a 98% Conf. Int. with a sample of 12?
2. When would you use a ‘t-distribution’ instead of a ‘z’?
What is the p-value for a t ≥ 2.03 with a sample size of 57?
Find the critical value ‘t’ that has 0.01 area to the right with a sample of 15.
What is the Margin of Error for a 98% Confidence Int. with a sample of 12 and a s = 0.658?
t-chart with 98%, 11 =
If the population standard deviation is unknown.
tcdf(2.03, e99, 56) = 0.024
t* = 2.624 =
Margin of Error =

2. WARM – UP (2)
A traffic light is yellow for 1.2 seconds. There is a suspicion that it is yellow much longer. What would the Type I AND the Type II errors be in this situation.
Conditions / Assumptions
Confidence Intervals – Formula, Interpretation, Calculation
9. P-value definition in context
10. T-test.
I – You feel that the light does last longer than 1.2 seconds, but it does NOT.
II – You feel that the light doesn’t last longer than 1.2 seconds, but it does.

3. WARM - UP
According to the Tourism Center of Buffalo, NY an average of 93” of snow falls per year. Having shoveled snow to unbury your car many times for 20 years, you suspect that the average is very different. What can you conclude from an SRS of 8 Buffalo winters?
78” 130” 140” 120” 108” 120” 156” 101”
μ = The true mean amount of Snow fall in Buffalo, NY yearly.
H0: μ = 93
Ha: μ ≠ 93
One Sample
t – Test
Since the P-Value is less than α = we REJECT H0 . There is evidence that Buffalo’s yearly snowfall average is NOT 93”.
SRS – Stated
Approximately Normal Distribution – Graph

4. t – Confidence Interval
EXAMPLE 1
Having shoveled snow to unbury my car many times for 20 years, I want to estimate the average amount of snow fall in Buffalo, NY. Use an SRS of 8 Buffalo winters to construct a 95% Confidence Interval:
78” 130” 140” 120” 108” 120” 156” 101”
μ = The true mean amount of Snow fall in Buffalo, NY yearly.
One Sample
t – Confidence Interval
We can be 95% Confident that the true mean amount of snowfall in Buffalo, NY will be between ” and ”
SRS – Stated
Approximately Normal Distribution – Graph

5. The Relationship between Confidence Intervals and Two-Sided Hypothesis Tests.
A level, α, two-sided (≠) significance test will reject the hypothesis H0: μ = μ0 exactly when the (1 – α) confidence interval for μ does not contain the true value μ0 .
Fail to Reject Region
Rejection Region
Rejection Region
EX: α = can be test with a 95% Confidence Interval

6. AFTER PARTY 4.2 H0: σ = 0.05 Ha: σ > 0.05
#3(b) To determine whether this sample of four bottles provides convincing evidence that the standard deviation of the amount of soda dispensed is greater than 0.05 ounce, a simulation study was performed. In the simulation study, 300 samples, each of size 4, were randomly generated from a normal population with a mean of 12.1 and a standard deviation of The sample standard deviation was computed for each of the 300 samples. The dotplot below displays the values of the sample standard deviations.
Use the results of this simulation study to explain why you think the sample provides or does not provide evidence that the standard deviation of the soda dispensed exceeds 0.05 fluid ounce.
s=0.085
H0: σ = 0.05
Ha: σ > 0.05
σ = The true standard deviation of soda amounts.
Since the P-Value is less than α = we REJECT H0 . There is evidence that standard deviation is > 0.05 ounces.

7. QUIZ REVIEW
Differences/Similarities between t and z (normal) distributions.
T values -> P-Values via tcdf(t, E99, df)
P values -> T-Values via InvT(prob/area to left, df)
Contextual definition of P-Value
Verifying Condition for T inferences
Type I / Type II Errors
Calculating Margin of Error via
Calculating Confidence Intervals via
T – Test via PHANTOMS (Stat -Test #2)
T – Intervals via PANIC (Stat -Test #8)

8. HW: Page 544: 27, 28, 33, 34

11. It has been reported that Louisiana Teachers make an average of $40,476 a year. You feel that teachers in the DFW area make more than this. The random sampling of 24 teachers in the area reveals an average of $42,410. (s = $4,580)
1. State the Null and Alternative Hypothesis.
2. Draw the Sampling Distribution and shade the probability representing the statistic.
H0: μ = 40476
Ha: μ > 40476
42410

12. Decision based on sample
STATISTICAL ERRORS (AGAIN)
Decision based on sample
Reject H0
Fail to Reject H0
H0 is True
H0 is False
Type I Error
POWER
Type II Error
Correct Decision
POWER = Probability of the test to correctly reject a false μ0. Increasing the Sample Size, n, or α (sign. Level) increases the Power of a test by decreasing Type II error. Power = 1 - Type II

13. List the Hypothesis and then Describe the Type I and Type II Error and the Consequences
1. In 2003 the Dept. of Commerce reported that the average American home cost $104,000. A recent slump in the US economy has greatly affected the housing market. You suspect that home prices have significantly decreased. If they have decreased you will buy a home.
H0: μ =
Ha: μ <
TYPE I ERROR = You feel that home prices have dropped, but in fact they remain the same. You buy a home and pay the high price.
TYPE II ERROR = You feel that home prices have remained the same, but in fact they have decreased. You don’t buy a home and lose out on bargain home prices.

14. If an error was made what type of error could have been made?
WARM - UP
Since the P-Value is less than α = we REJECT H0 . There is evidence Buffalo’s yearly snowfall average is NOT 93”.
If an error was made what type of error could have been made?
Describe the other type of error (were you fail to reject H0) in context to this example.
Type I
Type II - You feel that Buffalo’s yearly snowfall is 93” when in fact it is NOT.

15. Power – The Probability that a significance test will correctly reject a false H0.
Power = 1 – β (Type II Prob.)
Increasing sample sizes (n) will decrease Type II error (β) and consequently increase the Power of the test.
Increasing α = (Type I Prob.) will decrease β =(Type II Prob.).
Increasing α = (Type I Prob.) will increase the power. β α
μ μa

16. EXAMPLE: A credit card company has found that account holders use their cards an average of 8.4 times per month. To increase usage the company conducted a promotion for one month in which prizes could be won by using the card. During this month a random sample of 38 accounts were collected. The mean was 8.7 times with s = 2.6.
Is there sufficient evidence at the 0.05 level that the promotion increased usage?
μ = The true mean number of times account holders use
their cards per month.
H0: μ = 8.4
Ha: μ > 8.4
Since the P-Value > 0.05 the data IS NOT significant and there is NO evidence to REJECT H0 . The Promotion did NOT alter usage.
SRS – Stated
Approximately Normal Distribution – Since n is Large this is
true by the Central Limit Theorem.