Lesson SE4-SE5
SE4-SE5 Sources of association and Yates' correction Quiz: AQA Further Maths, Unit 4
20 questions
In partnership with Revision Ninja
Lesson SE4-SE5, Sources of association and Yates' correction: 20 multiple choice questions for the AQA Further Maths (7367), Unit 4: Optional application 2: statistics, written with Revision Ninja.
Host it live on the board and students join with a game code on their own devices, or revise alone with Free Play. The answers are revealed in the game.
The 20 questions
-
When is Yates' correction normally applied in a chi squared test?
- Only when the expected frequencies are all above 20
- Only when the null hypothesis is rejected
- To 2 x 2 tables with one degree of freedom
- To all tables with more than 5 rows
-
Which expression gives the chi squared statistic with Yates' correction?
- Sum of (|O - E| - 0.5)^2 / E
- Sum of (O - E - 0.5)^2 / E
- Sum of (O - E)^2 / (E - 0.5)
- Sum of (|O - E|)^2 / (E + 0.5)
-
What is the main purpose of Yates' correction?
- To improve the chi squared approximation for 2 x 2 tables
- To remove the need for expected frequencies
- To increase the value of the statistic
- To change the degrees of freedom to two
-
How are sources of association identified in a contingency table?
- By counting the number of rows
- By finding the cell with the smallest expected frequency
- By finding the largest observed frequency only
- By finding the cells where (O - E)^2 / E is largest
-
In a cell of a contingency table, what does an observed frequency greater than the expected frequency indicate?
- More observations occur in that combination than expected
- The test has failed
- Fewer observations occur in that combination than expected
- The variables are definitely independent
-
Which statement about association is correct?
- Association always proves causation
- Association does not imply causation
- Association means the variables are independent
- Association can only be found in 2 x 2 tables
-
What is the 1% critical value for a chi squared test with 1 degree of freedom?
- 6.635
- 3.841
- 5.991
- 9.210
-
A 2 x 2 table has observed frequencies 30, 10 in row 1 and 20, 40 in row 2. The uncorrected chi squared statistic is 16.67. At the 5% level with 1 degree of freedom, what is the conclusion?
- Reject H1, there is no association
- Accept H1 since the expected frequencies are 20 and 30
- Do not reject H0, there is no association
- Reject H0, there is evidence of association
-
For the 2 x 2 table with observed frequencies 20, 30 in row 1 and 30, 20 in row 2, the Yates-corrected statistic is 3.24. At the 5% level with 1 degree of freedom, what is the conclusion?
- Reject H0 since 3.24 is greater than 3.841
- Do not reject H0 since 3.24 is less than 3.841
- Reject H1 since 3.24 is less than 3.841
- Reject H0 since 3.24 is positive
-
A cell has observed frequency 30 and expected frequency 20. What is its contribution (O - E)^2 / E to the chi squared statistic?
- 10
- 2
- 5
- 1.5
-
For the same cell with observed 30 and expected 20, what is the Yates-corrected contribution, to 2 decimal places?
- 4.51
- 10
- 5
- 2.5
-
Compared with the uncorrected statistic, the Yates-corrected statistic is:
- Negative
- Larger
- Smaller
- Always equal
-
A 2 x 2 contingency table test gives a chi squared statistic of 6.2 with 1 degree of freedom. Is the result significant at the 5% level, given the critical value 3.841?
- Yes, since 6.2 is less than 3.841
- No, since 6.2 is greater than 3.841
- Yes, since 6.2 is greater than 3.841
- No, since 6.2 is less than 3.841
-
For a cell with observed frequency 12 and expected frequency 10, what is the Yates term (|O - E| - 0.5)^2 / E, to 3 decimal places?
- 0.225
- 0.2
- 0.4
- 0.9
-
In Yates' correction, what value is subtracted from |O - E| before squaring?
- 2
- 1
- 0.5
- 0.25
-
For the 2 x 2 table with observed frequencies 30, 10 in row 1 and 20, 40 in row 2, what is the Yates-corrected chi squared statistic, to 2 decimal places?
- 15.04
- 3.01
- 16.67
- 4.51
-
A 2 x 2 test gives Yates-corrected statistic 3.24 with 1 degree of freedom. What is the conclusion at the 5% significance level?
- No significant association at 5%
- Significant association at 1%
- The test cannot be carried out
- Significant association at 5%
-
A survey finds that more smokers than expected have lung disease, with a significant association in the test. Which conclusion is best supported?
- There is a positive association, but this alone does not prove smoking causes disease
- Smoking definitely causes the disease
- There is no association between the two
- The chi squared test is invalid in this context
-
An uncorrected chi squared statistic of 16.67 is compared with the 1% critical value for 1 degree of freedom. What is the conclusion?
- Not significant, since 16.67 is less than 6.635
- Significant at 1%, since 16.67 is greater than 6.635
- Significant at 1%, since 6.635 is greater than 16.67
- Not significant, since the degrees of freedom are 2
-
A cell has observed frequency 5 and expected frequency 8. What is its Yates-corrected term, to 2 decimal places?
- 0.94
- 0.31
- 0.78
- 2.25
Related quizzes
- Discrete random variables and their measures Quiz · SA1-SA3 · 20 questions
- Expectation and linear functions Quiz · SA4-SA5 · 20 questions
- Discrete uniform distribution Quiz · SA6-SA7 · 20 questions
- Modelling with the Poisson distribution Quiz · SB1-SB2 · 20 questions
- Mean, variance and sums of Poisson distributions Quiz · SB3-SB4 · 20 questions
- Hypothesis test for a Poisson mean Quiz · SB5 · 20 questions
- Type I and Type II errors and the power of a test · SC1-SC2 · 20 questions
- Probability density functions, medians and quartiles Quiz · SD1-SD3 · 20 questions
- Mean, variance and linear functions of continuous variables Quiz · SD4-SD5 · 20 questions
- Cumulative distribution functions and the rectangular distribution Quiz · SD6-SD8 · 20 questions