## Problem Description:

For question one, the homework deals with the analysis of a dataset that aims to understand the influence of variables on successful task completion. Specifically, it examines the interaction effect between two variables, T (representing training program) and X (representing motivation score) on the likelihood of successful task completion (Y). The goal is to determine whether the interaction between T and X is statistically significant and whether X alone is a significant predictor within different training programs.

For question two, the logistic regression analysis homework focuses on predicting the likelihood of a patient's death in the ICU based on their age. The logistic regression model is explored, including parameter estimation, significance tests, and confidence intervals for the model's coefficients. The goal is to assess the relationship between age and the likelihood of death in the ICU and provide a detailed interpretation of the results.

## Solution

### Question One: Interaction Analysis

In this section, we analyze the influence of variables T (training program) and X (motivation score) on the likelihood of successful task completion (Y). We investigate whether there is a statistically significant interaction effect between T and X and whether X alone predicts task completion within different training programs.

**Step 1: Assessing Interaction Effects
**

- Variables in the Equation:

Variable |
B |
S.E. |
Wald |
df |
Sig. |
Exp(B) |
95% C.I. for EXP(B) |
---|---|---|---|---|---|---|---|

Step 1a |
X |
-1.027 |
0.528 |
3 |
0.052 |
0.358 |
0.127 - 1.008 |

T(1) |
-38.195 |
16.053 |
5.661 |
1 |
0.017 |
0.000 |
0.000 - 0.001 |

T(1) by X |
2.498 |
1.036 |
5.815 |
1 |
0.016 |
12.157 |
1.596 - 92.587 |

Constant |
15.754 |
8.265 |
3.633 |
1 |
0.057 |
6950943.135 |

**a. Variable(s) entered on step 1: T * X.
**

The p-value of the Wald chi-square test for the interaction between T and X is 0.016, indicating a statistically significant interaction.

**Step 2: Simple Effect Analyses
**

We perform simple effect analyses by holding the effect of T constant and splitting the cases to observe the predicted variable estimates within Training Programs 1 and 2.

**Training Program 1:**

Variable |
B |
S.E. |
Wald |
df |
Sig. |
Exp(B) |
---|---|---|---|---|---|---|

Step 1a |
X |
1.471 |
0.891 |
2 |
0.099 |
4.354 |

Constant |
-22.441 |
13.762 |
2 |
0.103 |
0.000 |

The p-value for the effect of X on Y within Training Program 1 is 0.099.

**Training Program 2:**

Variable |
B |
S.E. |
Wald |
df |
Sig. |
Exp(B) |
---|---|---|---|---|---|---|

Step 1a |
X |
-1.027 |
0.528 |
3 |
0.052 |
0.358 |

Constant |
15.754 |
8.265 |
3 |
0.057 |
6950942.560 |

The p-value for the effect of X on Y within Training Program 2 is 0.052.

**Likelihood Ratio Test:
**

Omnibus Tests of Model Coefficients:

- Step 1 Chi-square: 26.668 (df = 1, p = 0.000)
- Training Program 1 Chi-square: 14.982 (df = 1, p = 0.000)
- Training Program 2 Chi-square: 11.809 (df = 1, p = 0.001)

The Chi-Square statistic for interaction (X and T) is 26.668. The likelihood ratio tests for the effects of X on Y within Training Programs 1 and 2 are 14.982 and 11.809, respectively.

The Chi-Square statistic for X by T(1) (entered after X) is 26.668 - 14.982 = 11.686 with a p-value of 0.00063, indicating a statistically significant improvement.

Consequently, it is clear that there is a statistically significant improvement after X is entered. The p-value is less than 0.05.

### Summary:

In this analysis, we observed a statistically significant interaction between T and X. The effects of X on Y within Training Programs 1 and 2 were not statistically significant. However, when X was entered after T(1), there was a statistically significant improvement.

### Question Two: Logistic Regression Analysis

For question two, the homework focuses on conducting a logistic regression analysis to predict the likelihood of a patient's death in the ICU based on their age. The logistic regression model is explored, and significance tests and confidence intervals for model coefficients are provided.

**Step 1: Data Exploration
**

We start by examining the frequency distribution of age groups and the proportion of patients who lived or died in each group.

Age Group |
STA Lived = 0 (Frequency) |
STA Dead = 1 (Frequency) |
Midpoint |
Proportion |
---|---|---|---|---|

15 – 24 |
24 |
2 |
19.5 |
0.0769 |

25 – 34 |
8 |
0 |
29.5 |
0 |

35 – 44 |
9 |
2 |
39.5 |
0.182 |

45 – 54 |
20 |
5 |
49.5 |
0.2 |

55 – 64 |
31 |
8 |
59.5 |
0.2052 |

65 – 74 |
41 |
9 |
69.5 |
0.18 |

75 – 84 |
21 |
9 |
79.5 |
0.3 |

85 – 94 |
6 |
5 |
89.5 |
0.455 |

**Step 2: Logistic Regression Model
**

- Variables in the Equation:

Variable |
B |
S.E. |
Wald |
df |
Sig. |
Exp(B) |
---|---|---|---|---|---|---|

Step 1a |
AGE |
0.028 |
0.011 |
1 |
0.009 |
1.028 |

Constant |
-3.059 |
0.696 |
1 |
0.000 |
0.047 |

**a. Variable(s) entered on step 1: AGE.
**

The logistic regression model:

**ln((p(x))/(1-p(x))) = -3.059 + 0.028x
**

The equation for the fitted values:

**p(x) = e^(-3.059 + 0.028x)/(1 + e^(-3.059 + 0.028x))
**

**Step 3: Significance Tests
**

- The P-value of the score test statistic for testing the significance of the slope coefficient for age is 0.007.
- The likelihood ratio test statistic value is 7.855 with a P-value of 0.005.
- The Wald test statistic value for testing the significance of the slope coefficient for age is 6.797 with a P-value of 0.009.

At an alpha level of 0.05, all p-values are less than 0.05, indicating that the tests are consistent with each other.

**Step 4: Deviance and Confidence Intervals
**

The deviance (i.e., -2 log-likelihood) of the fitted simple logistic regression model is 192.306.

**Step 5: Confidence Intervals
**

The 95% confidence interval for the slope coefficient for AGE is (0.006, 0.05).

This means that with 95% confidence, the population-level coefficient for age could range from 0.006 to 0.05.

**Step 6: Estimated Logistic Probability
**

- The estimated logit for a 60-year-old patient: g(60) = -1.379
- The estimated logistic probability for a 60-year-old patient: P(60) = 0.201

**Step 7: Confidence Interval for Estimated Probability
**

The 95% confidence interval for the estimated logistic probability of death in the ICU for a 60-year-old patient is (0.151, 0.263). This means that we are 95% confident that the probability of death for a 60-year-old patient falls within this range.

### Summary:

In this logistic regression analysis, we found that age is a significant predictor of ICU patient mortality. The 95% confidence interval for the age coefficient suggests that it is unequal to zero, and the estimated probability of death for a 60-year-old patient is 0.201 with a 95% confidence interval of (0.151, 0.263).