BES- Cheat- Sheet - Lecture notes 1-11 PDF

Preview

Summary

First year BES cheat sheet. Very helpful ;)...

Description

TOPIC 10 CONT: EXAMPLE: The mean consumption expenditure for a random sample of 45 households from a certain suburb is $270 p/w with standard dev of $30. Find and interpret 99% CI. t-critical for 99% and n=25 look in 99% column for df=45-1=44 t-critical=2.690 (tables) n=45, y-bar= $270p/w, s=$30p/w. CHECK CONDITIONS: Eqn =

270 ±

2.69∗30 =$ 270 ± $ 1 √ 45

= ($257 97 $282 030)

TOPIC 11 HYPOTHESIS TESTS ABOUT 1 PROPORTION/MEAN: Hypothesis testing general steps: 1. Write hypotheses 2. Check the conditions 3. Calculate the test statistic (uses formula) 4. Determine the rejection region (sketch) 5. Make a decision 6. Give a conclusion 1. HYPOTHESIS:

H0

Null hypothesis:

: express our

question into a statement about the population parameter, where we assume H0 is true

HA

Alternate Hypothesis:

: express our

claim into a statement about the population parameter that we are trying to PROVE is now true One tailed test: Eg. Ho:

μ

= 2, Ha:

μ

=2, Ha:

μ

>/< 2

μ

≠2

2. CONDITIONS tells us what type of model to use & type of test Eg. Conditions to test one proportion: - randomness – 10% condition –success/failure Eg. Conditions to test one mean: - randomness -10% condition –nearly normal 3. USE THE RELEVANT FORMULA Find the relevant calculated test statistic. testing Ho v Ha

Z-Calc =

^^p − p z= pq - for values of p n ´y −μ t-calc= s √n

√

4. SKETCH REJECTION REGION

individuals per cell (

≥5 ¿ .

Exp cell freq =

row total∗columntotal grand total X 2 calc=∑ , or sum of all baby chi’s:

→

( Obs − Expect Expected 2 (O−E ) E

Find

X2

critical and make decision

Standardised Residual with Chi-Squared:

Obs−exp √ exp

, if +, obs > exp, if –

obs < exp Unusual difference if beyond +2 & -2. - Determine the rejection region/sketch -Make a decision - give a conclusion

α = level of significance, 0.01 ≥ α ≥ 0.10

Here, use either Z or T tables to decide 5. MAKE A DECISION:

α

α

or

.

TOPIC 13: INFERENCE IN REGRESSION

6. GIVE CONCLUSION: tie decision to question **EXAMPLE: a TV company believes that more than 70% of TV viewers recognise its company logo. A random sample is taken & of the 49 sampled, 41 do recognise the logo. Test the claim at the 1% level of significance. n=49, p-hat=41/49, alpha=1%=0.01 i) critical value method:**** Ho: p = 0.70, Ha: p > 0.70 CHECK CONDITIONS:

Testing one mean uses a t-test

Formula to test one mean

´y −μ s √n

,

df=n-1 Example 2: A sample of 51 Australians revealed an average expenditure online of $352 with a standard dev of $95. We want to test whether the mean is significantly different from $326. Test using alpha 5%. Conditions: 1. Not told random, assume random but proceed with caution, 2. 51 < 10% of all AUS who shop online, 3. Nearly normal, n=51 >40, Y~N by CLT. alpha=5% in two tails, 2.5% in each tail. df= n-1=51-1=50 **use tables to find t-critical of

± 2.009

This example: Ho: beta1=0, Ha: beta10 therefore, 0.0167/2=0.0084 < any

α Reject Ho & accept Ha, ie. slope on dummy is significantly greater than 0 3. Interpret the 95% CI of wages

→

read straight off excel (-

0.299,-0.118) 4. R^2 = coefficient of determination How ‘good’ prediction is % of variation

TOPIC 17: TIME SERIES r= multiple R on excel, r^2= r square on excel

TOPIC 14: UNDERSTANDING RESIDUALS Point is unusual: large residual: if y value is far from y bar High leverage: x value far from majority of x vals Influential: removing the point, changes the regression equation & changes correlation *** Point is influential if changes regression eqn. TOPIC 15: MULTIPLE REGRESSION

for each cell inside table.

Critical Value for rej region: table, degrees of freedom df = (r-1) x (c-1), told alpha.

Alpha,

Either reject Ho and accept Ha at

Examples of inference, using MLR: 1. test whole eqn: Ho: betaW=BetaS=0 Ha: Betaw ≠ 0 and/or BetaS ≠ 0

→ Recall prob rules: A,B are independent: Pr(A&B) = Pr(A) x Pr(B) eg. poor attendance and pass Pr(PoorA&Pass) = P(poor) x P(pass) =21/55 x 29/55 = 0.201 (if independent) Therefore, we would expect 0.201 x 55 = 11.073 students to have poor att. & pass if independent Observed value = 4 Solution of tutorial attendance: Ho: grade & att. Are independent Ha: Grade & att. Are NOT independent CONDITIONS: 1. Counted Data Condition: the data must be counts for the categories of cat. Variables 2. Randomisation condition: the counted individuals should be a random sample of the population. 3. Expected cell frequency: expect at leasr 5

Direction is in Ha**

retain Ho at

Interpreting excel output: Standard Error= SE of residuals X-VALS: Standard Error: intercept – SE bo, wage – SE for b1. T-stat: t-calc test statistic for 2 tailed test about 0 P-Value: p-values for default 2 tailed test about 0 Lower/Upper95%: 95% CI for beta0 and beta1 CI FOR SLOPE (beta1) – read off excel Use CI formula for other %CI

2. Test dummy slope on dummy (shift) Ho: BetaS=0, BetaS>0

Calculated Chi-square statistic:

Two tailed test: Eg. Ho:

TOPIC 12: HYPOTHESIS TESTS FOR COUNTS Tests of independence: tests for evidence of association between two categorical variables Chi-squared tests we have the actual counts (observed) and have some way of saying what we would expect. Two Cat variables:

Quarterly Data: MARCH QTR: Q1: Jan, Feb, March JUNE QTR: Q2: April, May, June SEPT QTR: Q3: July, Aug, Sept DEC QRT: Q4: Oct, Nov, Dec Components of classical time series: 1. TREND: overall general movement over whole time series. Inc/dec/ flat?? 2. Cyclical Fluctuation; >1 year; repeated patterns, hard to see bc need lots of data, regular cycles with periods more than a year 3. Seasonal Variation;...