Posted: December 29th, 2022

**Application of the Pearson Correlation Coefficient and the Chi-Square Test.**

The purpose of this assignment is to practice calculating and interpreting the Pearson correlation coefficient and a chi-square test of independence. For this assignment, complete Problems 13.132 and 15.88 in the textbook. Include your process for conducting the calculations. You can complete the calculations by hand or using Excel or SPSS. If you use Excel or SPSS, copy and paste your output results into a Word document. When addressing each textbook problem, provide a response for each of the six steps of hypothesis testing listed below. Pick a test. Check the assumptions.

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Application of the Pearson Correlation Coefficient and the Chi-Square Test. List the hypotheses. Set the decision rule. Calculate the test statistic. Interpret the results. (What was done? What was found? What does it mean? What suggestions exist for future research?) Submit a Word document with your problem answers to each of the six steps. If Excel or SPSS was used to complete the assignment, submit the second Word document containing the screenshots to the instructor. This assignment uses a rubric. Please review the rubric prior to beginning the assignment to become familiar with the expectations for successful completion. You are not required to submit this assignment to LopesWrite.Please foloww the instruction very well.Application of the Pearson Correlation Coefficient and the Chi-Square Test.

Pearson correlation coefficient and chi-square

**Problem 13.132: Pearson correlation coefficient**

A sociologist wanted to see if there was a relationship between a family’s educational status and the eliteness of the college that there oldest child attended. She measured the educational status by counting how many years of education beyond high school the parents received. In addition, she measured the lateness of the school by its yearly tuition, in thousands (e.g., 5 = $5,000). She obtained a random sample of 10 families.

Years post-HS education | Yearly tuition ($) | |

1 | 0 | 12,000 |

2 | 7 | 26,000 |

3 | 8 | 33,000 |

4 | 8 | 18,000 |

5 | 4 | 20,000 |

6 | 5 | 7,000 |

7 | 12 | 15,000 |

8 | 17 | 38,000 |

9 | 8 | 41,000 |

10 | 2 | 5,000 |

The test statistic to determine the relationship between the variables has been identified as Pearson correlation coefficient test. The statistic meets the three assumptions for the test. Firstly, each variable is continuous. Secondly, each participant has a pair of values. Thirdly, there are no outliers in either variable. Based on these assumptions, the Pearson correlation coefficient test was conducted to was a relationship between a family’s educational status and the eliteness of the college that there oldest child attended (Goos&Meintrup, 2016).Application of the Pearson Correlation Coefficient and the Chi-Square Test.

The Pearson correlation coefficient was calculated using the Excel functions where Years post-HS education is the set of independent variables (array 1) and yearly tuition is the set of dependent variables (array 2). The Excel syntax for the function is presented as:

=PEARSON(array 1, array 2)

=PEARSON(b2:b11, c2:c11)

=__0.616869__

The result offered is __0.6169__, indicating a large/strong positive correlation between the two variables under review. The results suggest that a higher family’s educational status will be matched by a higher eliteness of the college that there oldest child attended, with the reverse being equally true (Goos&Meintrup, 2016).Application of the Pearson Correlation Coefficient and the Chi-Square Test.

**Problem 15.88: Chi-square**

A political scientist developed a theory that after an election, supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate. The day before a presidential election, he randomly selected parking lots, and at each selected parking lot, he randomly selected one care with a bumper sticker and recorded which candidate it supported. The day after the election, he followed the same procedure with a new sample of randomly selected parking lots. For both days, he then classified the bumper stickers as supporting the winning or losing candidate. Below are the results. Use hypothesis testing to see if a difference exists between how winners and losers behave.Application of the Pearson Correlation Coefficient and the Chi-Square Test.

Observed frequencies | Total | ||

Winner | Loser | ||

Before (observed)
Expected O-E (O-E) (O-E) |
34
39.3 -5.3 28.09 0.71 |
32
26.7 5.3 28.09 1.05 |
66
66 0
1.76 |

After (observed)
Expected O-E (O-E) (O-E) |
28
22.7 5.3 28.09 1.24 |
10
15.3 -5.3 28.09 1.84 |
38
38 0
3.08 |

Total | 62 | 42 | 104 |

The theory developed by the political scientist was tested using a chi-square test. This is a statistical test for independence by comparing two variables in a contingency table to determine whether they are related. The two variables are winners and loser with data collected for the period before and after the election. The presented data meets the six conditions for a chi-square test. Firstly, the data is presented in counts or frequencies. Secondly, data categories are mutually exclusive. Thirdly, a subject can only population one cell at every level of the variable. Fourthly, the two groups are independent. Fourthly, there are two categories of data offered as frequencies. Finally, the expected values in each cell exceeded five (McHugh, 2013).Application of the Pearson Correlation Coefficient and the Chi-Square Test.

The hypothesis presented is that supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate after an election.

Chi-square calculated value = 1.76+3.08 = __4.84__

Degrees of freedom = (r-1)(c-1) = (2-1)(2-1) = 1

Chi-square critical value for a p-value 0f 0.05= __3.84__

Given that the calculated chi-square value is greater than the critical chi-square values for a p-value of 0.05 (4.84>3.84, *p*=0.05), then we conclude that the data does not fit the model and reject the null hypothesis that supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate after an election (Goos&Meintrup, 2016).Application of the Pearson Correlation Coefficient and the Chi-Square Test.

**Problem 13.132: Pearson correlation coefficient**

A sociologist wanted to see if there was a relationship between a family’s educational status and the eliteness of the college that there oldest child attended. She measured the educational status by counting how many years of education beyond high school the parents received. In addition, she measured the lateness of the school by its yearly tuition, in thousands (e.g., 5 = $5,000). She obtained a random sample of 10 families.Application of the Pearson Correlation Coefficient and the Chi-Square Test.

Years post-HS education | Yearly tuition ($) | |

1 | 0 | 12,000 |

2 | 7 | 26,000 |

3 | 8 | 33,000 |

4 | 8 | 18,000 |

5 | 4 | 20,000 |

6 | 5 | 7,000 |

7 | 12 | 15,000 |

8 | 17 | 38,000 |

9 | 8 | 41,000 |

10 | 2 | 5,000 |

The Pearson correlation coefficient was calculated using the Excel functions where Years post-HS education is the set of independent variables (array 1) and yearly tuition is the set of dependent variables (array 2). The Excel syntax for the function is presented as:Application of the Pearson Correlation Coefficient and the Chi-Square Test.

=PEARSON(array 1, array 2)

=PEARSON(b2:b11, c2:c11)

=__0.616869__

The result offered is __0.6169__, indicating a large/strong positive correlation between the two variables under review. The results suggest that a higher family’s educational status will be matched by a highereliteness of the college that there oldest child attended, with the reverse being equally true(Goos&Meintrup, 2016).Application of the Pearson Correlation Coefficient and the Chi-Square Test.

**Problem 15.88: Chi-square**

A political scientist developed a theory that after an election, supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate. The day before a presidential election, he randomly selected parking lots, and at each selected parking lot, he randomly selected one care with a bumper sticker and recorded which candidate it supported. The day after the election, he followed the same procedure with a new sample of randomly selected parking lots. For both days, he then classified the bumper stickers as supporting the winning or losing candidate. Below are the results. Use hypothesis testing to see if a difference exists between how winners and losers behave.Application of the Pearson Correlation Coefficient and the Chi-Square Test.

Observed frequencies | Total | ||

Winner | Loser | ||

Before (observed)
Expected O-E (O-E) (O-E) |
34
39.3 -5.3 28.09 0.71 |
32
26.7 5.3 28.09 1.05 |
66
66 0
1.76 |

After (observed)
Expected O-E (O-E) (O-E) |
28
22.7 5.3 28.09 1.24 |
10
15.3 -5.3 28.09 1.84 |
38
38 0
3.08 |

Total | 62 | 42 | 104 |

The hypothesis test is that supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate after an election.

Chi-square calculated value = 1.76+3.08 = __4.84__

Degrees of freedom = (r-1)(c-1) = (2-1)(2-1) = 1

Chi-square critical value for a p-value 0f 0.05= __3.84__

Given that the calculated chi-square value is greater than the critical chi-square values for a p-value of 0.05 (4.84>3.84, *p*=0.05), then we conclude that the data does not fit the model and reject the null hypothesis that supporters of the losing candidate removed the bumper stickers from their cars faster than did supporters of the winning candidate after an election (Goos&Meintrup, 2016).

Application of the Pearson Correlation Coefficient and the Chi-Square Test.

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