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Fill in the two-way table based on the information below, compute the appropriat

ID: 3243255 • Letter: F

Question

Fill in the two-way table based on the information below, compute the appropriate test statistic value, degrees of freedom, p-value, and make a conclusion!

A large study selects 27000 humans to see if they lied or not and report their gender.

25% of samples are males who lied, and 33% are males who never lied.

27% of samples are females who lied and 15% are females who never lied

Please type whole numbers (integers in the table below):

Based on the above table, compute the appropriate test statistic (correct to 2 decimal places).

Hint: Types of test statistic are z-stat, t-stat, Chi-sq_stat, F_stat, etc. So first decide on the appropriate test and then find the value of the test statistics.

The test statistic value is:

Based on the above test statistic value, the p-value (correct to two decimal places) is:

Lied Male Female Total Yes - Observed Yes - Expected No - Observed No-Expected Total 27000

Explanation / Answer

Solution

To fill-in the table, we simply multiply 27000 by the respective percentages given.

Lied

Male

Female

Total

Yes - Observed

6750

7290

14040

Yes - Expected

No - Observed

8910

4050

12960

No-Expected

Total

15660

11340

27000

This is an example of Contingency Chi-square Test for Independence.

Hypotheses:

Null Hypothesis: H0: The two attributes (namely telling/not-telling lies and gender) are independent.     Vs    Alternative: HA: H0 is false.

Test Statistic

2 = [i = 1.2; j = 1, 2]{(Oij – Eij)2/Eij}, where Oij and Eij are respectively the observed and expected frequencies of the ijth cell in the table.

Under H0, Eij = (ith row total x jth column total)/Grand Total

Calculations:

Eij

Lied

Male

Female

Total

Yes - Expected

8143.2

5896.8

14040

No-Expected

7516.8

5443.2

12960

Total

15660

11340

27000

[Check: row totals and column totals of Eij must be equal to the corresponding row totals and column totals of Oij]

{(Oij – Eij)2/Eij} are given in the table below:

Lied

Male

Female

Total

Yes

238.35915

329.1626374

567.5217886

No

258.22241

356.5928571

614.8152709

Total

496.58156

685.7554945

1182.337059

Distribution and p-value

Under H0, test statistics, 2 ~ 2 distribution with degrees of freedom = 1[(number of rows - 1) x (Number of columns)]

So, p-value = P(21 > 1182.34) = 4.2008E-259

Decision:

Since p-value is very small, much less than even 0.1%, H0 rejected/

Conclusion:

There is sufficient statistical evidence to suggest that whether a person tells lie or not is dependent on the gender of the person.

DONE

Lied

Male

Female

Total

Yes - Observed

6750

7290

14040

Yes - Expected

No - Observed

8910

4050

12960

No-Expected

Total

15660

11340

27000

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