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36 questions · 1 formulas

p-value rule

p < α → reject H0

Trap: Fail to reject ≠ H0 proven true.

Hypothesis TestingE-level★ High priority — official Canvas example

Size p-value = 0.003. At 5%, reject or fail to reject H0: βSize=0?

→ A) Reject H0, because 0.003 < 0.05.

Canvas Example MCQ — Q63

  • A) Reject H0, because 0.003 < 0.05.
  • B) Fail to reject H0, because 0.003 < 0.05.
  • C) Reject H0, because 0.003 > 0.05.
  • D) Fail to reject H0, because 0.003 > 0.05.

0.003 < 0.05, so the coefficient is statistically significant at 5%.

Rule: p<α -> reject.

Hypothesis TestingE-level★ High priority — official Canvas example

Age p-value = 0.425. At 5%, reject or fail to reject H0?

→ B) Fail to reject H0, because 0.425 > 0.05.

Canvas Example MCQ — Q64

  • A) Reject H0, because 0.425 > 0.05.
  • B) Fail to reject H0, because 0.425 > 0.05.
  • C) Reject H0 because the estimated coefficient is negative.
  • D) Reject H0, because 0.425 < 0.05.

0.425 > 0.05. There is insufficient evidence at 5% to reject βAge=0.

Rule: p>α -> fail to reject.

Hypothesis TestingE-level★ High priority — official Canvas example

Balcony p-value = 0.083. At 5%, reject or fail to reject?

→ C) Fail to reject H0, because 0.083 > 0.05.

Canvas Example MCQ — Q65

  • A) Reject H0, because 0.083 < 0.10.
  • B) Reject H0, because 0.083 > 0.05.
  • C) Fail to reject H0, because 0.083 > 0.05.
  • D) Fail to reject H0 because the coefficient is positive.

0.083 > 0.05. Balcony is not significant at the 5% level.

Rule: Sign depends on p-value, not coefficient sign.

Hypothesis TestingE-level★ High priority — official Canvas example

Same Balcony p-value 0.083, but use 10% significance.

→ A) Reject H0, because 0.083 < 0.10.

Canvas Example MCQ — Q66

  • A) Reject H0, because 0.083 < 0.10.
  • B) Fail to reject H0, because 0.083 < 0.10.
  • C) Reject H0, because 0.083 > 0.10.
  • D) Fail to reject H0, because 0.083 > 0.10.

0.083 < 0.10, so Balcony is significant at 10%.

Rule: A variable can be significant at 10% but not 5%.

Hypothesis TestingE-level★ High priority — official Canvas example

At 5%, Size p=.003, Age p=.425, Balcony p=.083. Which explanatory variables are significant?

→ A) Size only

Canvas Example MCQ — Q67

  • A) Size only
  • B) Size and Balcony
  • C) Age and Balcony
  • D) All three explanatory variables

Only 0.003 is below 0.05.

Rule: Compare each p-value to the same α.

Hypothesis TestingE-level★ High priority — official Canvas example

Which variable has p-value above 10%: Size .003, Age .425, Balcony .083?

→ B) Age

Canvas Example MCQ — Q68

  • A) Size
  • B) Age
  • C) Balcony
  • D) Size and Balcony

0.425 > 0.10; the other two are below 0.10.

Rule: Convert “above 10%” to p>0.10.

Hypothesis TestingE-level★ High priority — official Canvas example

Correct decision rule for a two-sided test at 5%?

→ B) Reject H0 if p < 0.05.

Canvas Example MCQ — Q69

  • A) Reject H0 if p > 0.05.
  • B) Reject H0 if p < 0.05.
  • C) Reject H0 only if the coefficient is positive.
  • D) Reject H0 only if the t-statistic is positive.

The p-value is the smallest significance level at which H0 would be rejected.

Rule: Reject when p is smaller than α.

Hypothesis TestingE-level★ High priority — official Canvas example

Age p=.425 at 5%. What does failing to reject mean?

→ C) We fail to reject H0.

Canvas Example MCQ — Q70

  • A) We reject H0.
  • B) We prove that beta_Age = 0.
  • C) We fail to reject H0.
  • D) We prove that age has no relationship with house prices.

A non-small p-value means insufficient evidence against H0, not proof that H0 is true.

Rule: “Fail to reject” is deliberately weaker than “accept/prove.”

ProbitD–A levelVery high priority

Is the coefficient of the NETWORTH variable significant at 2% level?

→ B) Yes

Old Exam — Oct 2023, Q15 · adapted

Given: Probit Model 1 output: NETWORTH coefficient 0.0864, t ≈ 2.44, p-value ≈ 0.017.

  • A) No
  • B) Yes
  • C) Yes because the p-value is higher than 2 %.
  • D) No because the p-value lower than 2 %.

p for NETWORTH is about 0.017. Since 0.017<0.02, significant at 2% -> B.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

ProbitD–A levelVery high priority

Is the coefficient of the NETWORTH variable significant at 1% level?

→ A) No

Old Exam — Oct 2022, Q21

Given: Probit Model 1 from course: NETWORTH coefficient≈0.0864, SE≈0.0354, so t≈2.44 and p≈0.017.

  • A) No
  • B) Yes
  • C) Yes because the p-value is higher than 1%
  • D) No because the p-value lower than 1 %

NETWORTH t≈0.0864/0.0354≈2.44, giving p≈0.017. Since 0.017>0.01, it is not significant at 1% -> A.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Panel DataD–A levelNormal priority

You want to estimate a panel data regression model. Your key variable of interest is time-invariant. Therefore, you should estimate a_______

→ B) Random effects model

Old Exam — Oct 2023, Q17

Given: Time-invariant key regressor: fixed effects wipes it out; random effects can estimate its coefficient.

  • A) Fixed effects model
  • B) Random effects model
  • C) A fixed effects (FE) model because time-invariant variables are not eliminated with FE model.

A time-invariant regressor has no within-person variation and is eliminated by FE. To estimate its coefficient, use RE (subject to RE assumptions) -> B.

Rule: Time-invariant variable -> FE cannot estimate its coefficient. Small Hausman p-value -> reject RE, prefer FE.

Panel DataD–A levelNormal priority

You have done a Hausman test to see if a random effects or a fixed effects models should be used. The p-value of the Hausman test is 0.0003. Therefore you conclude that ___________.

→ A) Fixed effects model should be used.

Old Exam — Oct 2023, Q18

Given: Hausman p=0.0003: reject RE consistency; prefer FE.

  • A) Fixed effects model should be used.
  • B) Random effects model should be used.

Hausman H0: RE is consistent. p=0.0003 is tiny -> reject H0 -> prefer FE -> A.

Rule: Time-invariant variable -> FE cannot estimate its coefficient. Small Hausman p-value -> reject RE, prefer FE.

Panel DataD–A levelNormal priority

In this course (AI2152) we have studied the NLS panel data set. Suppose that you want to estimate panel data models and that you have collected data on individuals with following variables: Variable definitions: Which one of the following statements is true?

→ B) You want to estimate a random effects model when you are interested the coefficient of the variable black.

Old Exam — Dec 2022, Q28

Given: Panel setup: BLACK is time-invariant for each person; UNION varies over time.

  • A) You want to estimate a fixed effects model when you are interested the coefficient of the variable black.
  • B) You want to estimate a random effects model when you are interested the coefficient of the variable black.
  • C) Both fixed and random effects model are appropriate when you are interested the coefficient of the variable black.
  • D) The variable union is a so-called time-invariant variable.
  • E) None of the above (A, B, C, D) is close to be correct.

Fixed effects eliminate variables that do not change over time within each individual. BLACK is time-invariant, so its coefficient cannot be estimated by FE; RE can estimate it -> B.

Rule: Time-invariant variable -> FE cannot estimate its coefficient. Small Hausman p-value -> reject RE, prefer FE.

Panel DataD–A levelNormal priority

How do you test for endogenous regressors, or correlation between the error term and any regressor in a panel data random effects model?

→ A) Estimate coefficients with RE and FE; then perform a Hausman test.

Old Exam — Oct 2021, Q24

Given: Hausman concept: estimate FE and RE; test whether differences in common coefficients are systematic.

  • A) Estimate coefficients with RE and FE; then perform a Hausman test.
  • B) Estimate the model capturing estimated residuals; then regress residuals on all regressors and perform an Chi2-test
  • C) Estimate RE model capturing estimated residuals; then estimate coefficients of correlation with each regressor.
  • D) Estimate RE and FE models and perform an F test on each model individually. If the difference between the F statistics is significant, conclude endogeneity.

Hausman compares FE and RE estimates. If differences are systematic, RE is inconsistent and FE is preferred -> A.

Rule: Time-invariant variable -> FE cannot estimate its coefficient. Small Hausman p-value -> reject RE, prefer FE.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, what is the null hypothesis of the chi-square test?

→ B) There is no relationship between the primary reason for not returning and the hotel.

Old Exam — Dec 2022, Q14

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

  • A) There is a relationship between the primary reason for not returning and the hotel.
  • B) There is no relationship between the primary reason for not returning and the hotel.
  • C) There is a positive relationship between the primary reason for not returning and the hotel.
  • D) There is a negative relationship between the primary reason for not returning and the hotel.
  • E) None of the above (A, B, C, D) is close to be correct.

Chi-square workflow: H0=no association/same proportions; E=(row total x column total)/grand total; statistic=sum((O-E)^2/E); df=(r-1) (c-1); reject if p<alpha or statistic>critical. Source answer: B) There is no relationship between the primary reason for not returning and the hotel..

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the value of the chi-square test statistic is closest to______.

→ B) 3.6018

Old Exam — Dec 2022, Q15 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

  • A) 4.2962
  • B) 3.6018
  • C) 0.7304
  • D) 27.4104
  • E) None of the above (A, B, C, D) is close to be correct.

For a contingency table, chi-square=sum((O-E)^2/E). Using the supplied 4x3 hotel table gives about 3.6018.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the upper-tail critical value of the test statistic at 1% significance level (α = 0.01) is closest to_________.

→ D) 16.8119

Old Exam — Dec 2022, Q16 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

  • A) 27.4104
  • B) 12.5916
  • C) 3.6018
  • D) 16.8119
  • E) None of the above (A, B, C, D) is close to be correct.

df=(rows-1)(columns-1)=(4-1)(3-1)=6. The 1% upper-tail chi-square critical value for df=6 is about 16.8119.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the conclusion for the Chi-square test above, at 5 percent significance level (α = 0.05), would be that ______.

→ B) the null hypothesis will not be rejected since the computed p-value is 0.7304.

Old Exam — Dec 2022, Q17 · adapted

Given: Hotel table (counts): Price 18/9/36; Location 23/15/42; Room 9/8/16; Other 12/14/25 across three hotels.

  • A) the null hypothesis will be rejected since the computed p-value is 0.0001.
  • B) the null hypothesis will not be rejected since the computed p-value is 0.7304.
  • C) the null hypothesis will be rejected since the computed p-value is 0.7304.
  • D) the null hypothesis will be rejected since the computed p-value is 16.8119.
  • E) None of the above (A, B, C, D) is close to be correct.

p≈0.7304>0.05, so fail to reject H0. Do not say that H0 is proved true.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, what is the null hypothesis of the chi-square test?

→ B) There are no differences among the three hotels in the proportion of guests who experienced the hotel visits as luxury.

Old Exam — Oct 2022, Q10

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

  • A) All hotel guests have experienced that the hotel visits were luxury.
  • B) There are no differences among the three hotels in the proportion of guests who experienced the hotel visits as luxury.
  • C) There are large differences among the three hotels in the proportion of guests who experienced the hotel visits as luxury.
  • D) Fifty percent of the hotel guests in all hotels experienced the hotel visits as luxury.
  • E) None of the above (A, B, C, D) is close to be correct.

For a chi-square homogeneity/independence test, H0 says the proportions/distributions are the same across groups - here, no hotel difference in the luxury proportion.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the value of the chi-square test statistic is closest to______.

→ A) 3.5967

Old Exam — Oct 2022, Q11 · adapted

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

  • A) 3.5967
  • B) 5.9915
  • C) 159.2251
  • D) 2.7165.
  • E) None of the above (A, B, C, D) is close to be correct.

Compute expected counts from row total x column total / grand total, then sum (O-E)^2/E. Source result ≈3.5967.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the conclusion for the Chi-square test above, with 5 percent significance level, would be that ______.

→ C) The p-value must be higher than 5 %, implying that the null hypothesis will not be rejected.

Old Exam — Oct 2022, Q12 · adapted

Given: Hotel table: 400 guests per hotel. Luxury yes counts 344,356,338; not-luxury 56,44,62.

  • A) The p-value must be lower than 5 %, implying that the null hypothesis will be rejected.
  • B) The p-value must be lower than 5 %, implying that the null hypothesis will not be rejected.
  • C) The p-value must be higher than 5 %, implying that the null hypothesis will not be rejected.
  • D) The p-value must be higher than 5 %, implying that the null hypothesis will be rejected.
  • E) None of the above (A, B, C, D) is close to be correct.

With df=2, chi-square≈3.5967 gives p>0.05. Therefore fail to reject H0.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the expected frequency of satisfied customers from the Los Angeles sample is ________.

→ A) 60

Old Exam — Dec 2021 re-exam, Q13

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300.

  • A) 60
  • B) 75
  • C) 80
  • D) 90
  • E) None of the above (A, B, C, D) is close to be correct.

Expected satisfied in Los Angeles = row total 225 x LA column total 80 / grand total 300 = 60 -> A.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the chi-square test statistic for these samples is ________.

→ B) 2.44

Old Exam — Dec 2021 re-exam, Q14 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300.

  • A) 1.49
  • B) 2.44
  • C) 4.15
  • D) 5.33
  • E) None of the above (A, B, C, D) is close to be correct.

Compute all expected counts and sum (O-E)^2/E; source result≈2.44 -> B.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the chi-square critical value using α = 0.5 is ________.

→ A) 5.991

Old Exam — Dec 2021 re-exam, Q15 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300. Source question prints alpha=0.5, but the keyed 5.991 critical value is the 5% (0.05) critical value for df=2. Treat 0.5 as a source typo.

  • A) 5.991
  • B) 3.841
  • C) 2.706
  • D) 7.815
  • E) None of the above (A, B, C, D) is close to be correct.

For df=(2-1)(3-1)=2 and alpha=0.05, chi-square critical≈5.991 -> A. The printed alpha=0.5 is inconsistent with the keyed answer and is treated as a typo.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, using α = 0.05, the conclusion for this chi-square test would be that because the test statistic is

→ C) less than the critical value, we cannot reject the null hypothesis and cannot conclude that there is a difference in proportion of satisfied customers between these three locations

Old Exam — Dec 2021 re-exam, Q16 · adapted

Given: Restaurant table: satisfied counts 80,85,60; sample sizes 100,120,80. Total satisfied=225 of 300. With chi-square≈2.44 and 5% critical≈5.991, do not reject H0.

  • A) more than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations
  • B) less than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations
  • C) less than the critical value, we cannot reject the null hypothesis and cannot conclude that there is a difference in proportion of satisfied customers between these three locations
  • D) more than the critical value, we cannot reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations. 6

2.44<5.991, so fail to reject H0; there is insufficient evidence of different satisfaction proportions -> C.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the upper-tail critical value from a chi-square distribution at the 0.01 level of significance is______.

→ A) 6.63.

Old Exam — Oct 2021, Q11

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88. df=1.

  • A) 6.63.
  • B) 5.99.
  • C) 20.74.
  • D) 2.71.
  • E) None of the above (A, B, C, D) is close to be correct.

The table is 2x2, so df=(2-1)(2-1)=1. The 1% upper-tail chi-square critical value is about 6.63 -> A.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the value of the chi-square test statistic is ______.

→ C) 20.74.

Old Exam — Oct 2021, Q12 · adapted

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88

  • A) 6.63.
  • B) 5.99.
  • C) 20.74.
  • D) 2.71.
  • E) None of the above (A, B, C, D) is close to be correct.

Using E=row total x column total / grand total and summing (O-E)^2/E gives chi-square≈20.74 -> C.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Chi-square & Time SeriesD–A levelLegacy / lower priority

Using the table, the conclusion for a Chi-square test would be that ______.

→ D) The p-value must be lower than 1 %, implying that the null hypothesis will be rejected.

Old Exam — Oct 2021, Q13 · adapted

Given: 2x2 table: Exclusive speakers: Like Dune=108, Not like=95; Not-exclusive: Like=33, Not like=88

  • A) The p-value must be higher than 10 %, implying that the null hypothesis will be rejected.
  • B) The p-value must be higher than 10 %, implying that the null hypothesis will not be rejected.
  • C) The p-value must be lower than 1 %, implying that the null hypothesis will not be rejected.
  • D) The p-value must be lower than 1 %, implying that the null hypothesis will be rejected.
  • E) None of the above (A, B, C, D) is close to be correct.

20.74 is far above the 1% critical value 6.63, so p<0.01 and H0 is rejected -> D.

Rule: Use expected counts, df=(r-1)(c-1), and p-value/critical-value decision rules.

Panel DataD–A levelNormal priority

If you perform a Hausman test on a random effects model and have a test statistic that exceeds your critical value, which of the following is NOT correct?

→ B) None of the common coefficient estimates in the RE model will be significantly different in the FE model.

Old Exam — Dec 2021 re-exam, Q29

  • A) At least one of the regressors in the RE model is endogenous.
  • B) None of the common coefficient estimates in the RE model will be significantly different in the FE model.
  • C) FE may be the preferred estimation technique.
  • D) This model may be better estimated using the Hausman-Taylor estimator.
  • E) None of the above (A, B, C, D) is correct.

Rejecting the Hausman H0 means RE and FE coefficients differ systematically. So B (no differences) is the statement that is NOT correct. A, C and D are all consistent with rejection.

Rule: Large Hausman statistic / small p-value → reject RE consistency → prefer FE.

⚠ The original source key marks D. The wording is ambiguous; B is the statement that clearly contradicts a Hausman rejection. Memorize the rule, not the letter.

Hypothesis TestingE-levelHigh priority

In Model 5, the p-value for calw_pct is 0.414578. At significance levels α = 0.1%, 1%, and 5%, should H0: β_calw = 0 be rejected?

→ A) Fail to reject H0 at all three levels

Computer Lab 4 — Exercise 3, Task 2b

  • A) Fail to reject H0 at all three levels
  • B) Reject H0 at all three levels
  • C) Reject only at 5%
  • D) Reject only at 0.1% and 1%

Since 0.414578 exceeds 0.1%, 1%, and 5% (all of which are far smaller than 0.4146), H0 is not rejected at any of the three levels.

Rule: Reject H0 only if p-value < α; otherwise fail to reject.

Hypothesis TestingE-levelHigh priority

In Model 5, meal_pct has a p-value of approximately 0.000000, and str has a p-value of 0.000161. At α = 0.1%, should H0 be rejected for both variables?

→ A) Yes, reject H0 for both, since both p-values are below 0.001

Computer Lab 4 — Exercise 3, Task 2b

  • A) Yes, reject H0 for both, since both p-values are below 0.001
  • B) No, fail to reject for both
  • C) Reject only for meal_pct
  • D) Reject only for str

Both 0.000000 and 0.000161 are less than α = 0.001 (0.1%), so H0 is rejected for both variables at this strict significance level.

Rule: Compare each p-value to α directly; reject when p-value < α.

Hypothesis TestingD–A levelHigh priority

With df = n − k − 1 = 420 − 4 − 1 = 415, the two-sided 5% critical value is |t| = 1.9657. The reported t-statistic for el_pct is -3.581. Should H0: β_el_pct = 0 be rejected at the 5% level?

→ A) Yes, because |-3.581| = 3.581 > 1.9657

Computer Lab 4 — Exercise 3, Task 2c

  • A) Yes, because |-3.581| = 3.581 > 1.9657
  • B) No, because 3.581 < 1.9657 is false reasoning so fail to reject
  • C) Cannot be determined without the p-value
  • D) No, because t-statistics cannot be used for two-sided tests

Reject H0 when |t| exceeds the critical value: |-3.581| = 3.581, which exceeds 1.9657, so H0 is rejected at the 5% level.

Rule: Two-sided rule: reject H0 if |t| > t_critical.

Hypothesis TestingE-levelHigh priority

For the same α and the same two-sided test specification, do the p-value method and the critical-value (t-statistic) method always produce the same reject/fail-to-reject decision?

→ A) True — they are equivalent decision rules for the same test

Computer Lab 4 — Exercise 3, Task 2d

  • A) True — they are equivalent decision rules for the same test
  • B) False — the t-statistic method is more conservative
  • C) False — the p-value method is only valid for one-sided tests
  • D) True, but only when the sample size exceeds 1000

Both approaches test the same null against the same alternative using the same α, so 'p-value < α' and '|t| > t_critical' always agree for a correctly specified two-sided test.

Rule: p-value method and critical-value method are mathematically equivalent given the same α and test.

Hypothesis TestingE-levelHigh priority

In a regression output, a coefficient is marked with ** (two stars). Based on the standard convention (*p<0.10, **p<0.05, ***p<0.01), what does this indicate?

→ A) The p-value for that coefficient is below 0.05

Computer Lab 4 — Exercise 3, Task 5

  • A) The p-value for that coefficient is below 0.05
  • B) The coefficient's economic magnitude is large
  • C) The coefficient explains 5% of the variance in Y
  • D) The p-value is below 0.01

Two stars (**) conventionally denote statistical significance at p < 0.05. More stars indicate stronger statistical evidence against H0, not larger economic magnitude.

Rule: Significance stars reflect p-value thresholds, not effect size or variance explained.

Regression BasicsE-levelVery high priority

Switching from Excel's standard OLS output to Python's HC1 robust standard errors for the same ln(PRICE) regression, which outputs should remain identical, and which can change?

→ A) Coefficients and fitted values stay the same; standard errors, t-statistics and p-values can differ

Computer Lab 5 — Exercise 4, Robust SE

  • A) Coefficients and fitted values stay the same; standard errors, t-statistics and p-values can differ
  • B) Everything, including coefficients, changes
  • C) Only the R² changes
  • D) Coefficients change but standard errors stay identical

OLS point estimates and fitted values depend only on the OLS fitting procedure, which is unaffected by the choice of variance estimator; only the estimated standard errors (and resulting t-stats/p-values) can change with HC1.

Rule: HC1 robust SEs change only the measure of estimation uncertainty, not the OLS coefficients themselves.

Hypothesis TestingE-level★ High priority — official Canvas example

A financial-return regression reports p-values: interest rate p = .009, recession dummy p = .089, credit spread p = .455. At the 10% level, which variables reject H0: β = 0?

→ C) Interest rate and Recession dummy

Canvas Example MCQ — Q72

  • A) Interest rate only
  • B) Recession dummy only
  • C) Interest rate and Recession dummy
  • D) Interest rate and Credit spread
  • E) All three explanatory variables

.009 < .10 and .089 < .10; .455 > .10.

Rule: Reject H0 when p < α; changing α can change the conclusion.