**Chi-Square**** Tests**

**Jaspreet | Prabhakaran | Tanya | Prashant**

**Chi-Square**** Tests**

**Jaspreet | Prabhakaran | Tanya | Prashant**

A mathematical comparison between expected frequencies and observed frequencies which are developed on the basis of two main hypotheses.

1) Null Hypothesis

2) Alternative Hypothesis

Chi-Square Tests

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A researcher plans to ask employees whether they favour,

oppose or are indifferent about a change in the company’s superannuation provider.

How can we formulate a null hypothesis for a Chi-Square test to determine

the expected frequency for each answer.

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Question 1

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**Answer 1**

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1) We assume that the Null Hypothesis is true

3) The expected frequency is an equal distribution i.e. 1/3 of 75 for each available response

2) We have chosen a random sample of 75 employees

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Suppose the observed responses to the proposed survey in

Q:1 were: **Favour** – 30, **Oppose** – 15, **Indifferent** – 25.

How can we perform a Chi-Square test based on these observed responses and the expected frequencies provided in the previous question?

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Question 2

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**Pvalue = P(χ2)****(Degree of freedom) = (3-1) = 2****Significance level – (10 %)**

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**Answer 2**

Calculation of Chi-Square Test

Outcomes of Chi-Square Test

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Determination of Significance Value (α)

- For d.f. of 2, 4.6 is the critical value.
- Because Pvalue = P(χ2 > .10) there is more than 10% probability of the expected outcomes to be true.
- Therefore the null hypothesis is not rejected.

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The following is a summary of typical types of data that might be received

from a questionnaire to discover the local populations agreement and

disagreement with the success of the new neighborhood

initiatives designed to enhance well being.

Determine a statistical hypothesis and perform a Chi-Square test on the data.

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Question 3

Public Transport Facilities in the Neighbourhood have improved over the last year.Public Transport Facilities in the Neighbourhood have improved over the last year.

1

Public Transport Facilities in the Neighborhood have improved over the last year.

Agree - 20

Neutral - 36

Disagree - 28

Total - 84

Scenario

2

Generally speaking, there is less rubbish lying around the neighborhood now than a year ago.

Agree - 48

Neutral - 22

Disagree - 12

Total - 82

Scenario

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**Answer 3**

**Pvalue = P(χ2)****(Degree of freedom) = (3-1) = 2**

Calculation of Chi-Square Test

Outcomes of Chi-Square Test

**Scenario 1**

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Determination of Significance Value (α)

- Significance level – (10 %)
- For d.f. of 2, 4.6 is the critical value.
- Because Pvalue = P(χ2 > .10) there is more than 10% probability of the expected outcomes to be true.
- Therefore the null hypothesis is not rejected.

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**Answer 3**

- Pvalue = P(χ2 < 0.01)
- (Degree of freedom) = (3-1) = 2
- Significance level (α) – 1 (%)

Calculation of Chi-Square Test

Outcomes of Chi-Square Test

**Scenario 2**

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Determination of Significance Value (α)

- Because χ2 < 1%, our null hypothesis will be rejected.
- Significance level (α) – 1 (%)
- Because χ2 < 1%, our null hypothesis will be rejected.

**DISCUSSION QUESTION**

**What are the statistical decisions that can be made if there is a change in the significance level?**

Thank you!