__Determination of nature of relationship between variables, One way, Two way ANOVA and Chi-Square test__

__Determination of nature of relationship between variables, One way, Two way ANOVA and Chi-Square test__

The questions asked here are related to statistics. In the question, examiner has asked the students to find out whether there is a linear relationship between two variable sets after deriving a regression line formula. In another question two sets of variables are given. Students are required to find out which variable set is influencing the conclusion to a higher degree. In one of the questions asked here, students are required to conduct a One-way ANOVA on the data and report the test statistics using correct APA formatting and interpret the results. In penultimate question, students have been asked to use the Chi-Square test for independence at a level of significance of 0.05. In the last question there is an application of Two-way analysis of variance (ANOVA) and reporting of results has been asked to be done using correct APA style.

** SOLUTION** : –

Descriptive Statistics |
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N | Minimum | Maximum | Mean | Std. Deviation | |

hours | 20 | 3.00 | 15.00 | 8.8500 | 3.66024 |

Valid N (listwise) | 20 |

the mean hours of exercise per week by the participants is 8.85

B

of the hours of exercise per week by the participants

Descriptive Statistics |
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N | Std. Deviation | Variance | |

hours | 20 | 3.66024 | 13.397 |

Valid N (listwise) | 20 |

Variance is of the hours of exercise per week by the participants13.397 and standard deviationof the hours of exercise per week by the participants is 3.66024

C

r=-0.103 and since it is close to 0, it indicates a weak, negative linear relationship between the hours in a week spent exercising and a person’s life satisfaction. In general the more hours in a week spent exercising the less the person’s life satisfaction.

D

The coefficient of hours is -0.070, this means that a unit increase in hours in a week spent exercising decreases a person’s life satisfaction by 0.070. t=-0.441,p=0.664. The p-value is 0.664>0.05, this means the effect of hours in a week spent exercising is not significant.

The R^{2} is 0.011. this means that 1.1% of variation in person’s life satisfaction is explained by hours in a week spent exercising

ANOVA^{a} |
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Model | Sum of Squares | df | Mean Square | F | Sig. | |

1 | Regression | 1.252 | 1 | 1.252 | .195 | .664^{b} |

Residual | 115.698 | 18 | 6.428 | |||

Total | 116.950 | 19 |

E
Life satisfaction ranking =5.761-0.070 hours in a week spent exercising |

2

Step 1: since both groups are made up of different people and we are comparing means, we will run a two-independent sample t-test.

Step 2: let u_{1}=mean of people taking pharmaceuticals and u_{2 }is the mean of those taking the new relaxation technique, we have:

H_{0}: u_{1}-u_{2}=0

H_{1}: u_{1}-u_{2}<0

Step 3

Step 4: From the SPSS output, we see our test statistic is t(44.595) = 2.482 with a p-value of 0.017/2 = 0.0085.

Step 5: since p-value is 0.0085<0.05, we reject H_{0}. Therefore, we do not have enough evidence to say the mean amount of time taken to fall asleep is the same for both the pharmaceutical and relaxation group. The mean amount of time taken to fall asleep for the pharmaceutical group is 102.3 while for treatment group is 74.7 and the difference is statistically significant. Therefore, we conclude that the new relaxation technique is more effective in treating insomnia than the pharmaceutical technique.

3

Step 1: Since we are comparing means of more than 2 independent groups, and since we only have one independent variable, life satisfaction rating, then we will use a one-way ANOVA.

Step 2: let u_{1} be the mean of life satisfaction rating for respondents who have no interaction with their mother, u_{2} =the mean of life satisfaction rating for respondents who have low interaction with their mother u_{3}= the mean of life satisfaction rating for respondents who have moderate interaction with their mother and u_{4}= the mean of life satisfaction rating for respondents who have high interaction with their mother, we have:

H_{0}:u_{1}=u_{2}=u_{3}=u_{4}

H_{1}: at least one group mean is different

Step 3:

Step 4

From the SPSS output, we see our test statistic is F(3, 36) = 1.570 with a p-value of 0.213.

Step 5

Since the p-value =0.213>0.05, we do not reject H_{0}. Therefore, we have no evidence to say there is a relationship between level of interaction a woman has with her parent and life satisfaction rating.

4

Step 1: Since we are looking at the relationship of two nominal variables, we will use a chi-square test of independence.

Step 2:

Ho: there is no relationship between handedness and gender

Ha: there is a relationship between handedness and gender

Step 3

Step 4

From the SPSS output, we see our test statistic is X² = 4.114 with a p-value of 0.043.

Step 5

Since the chi-square p-value is 0.043<0.05, we reject H_{o}. therefore, we have evidence to say there is relationship between handedness and gender. From the bar chart, we can tell that men are more likely to be right-handed and women are more likely to be left-handed.

5

Step 1: Since we are comparing means and we have two independent variables, diet and exercise status, we will use a two-way ANOVA test.

Step 2

H_{0}: mental acquity do not vary by diet, exercise or a combination of both

H_{a}: mental acquity do vary by diet, exercise or a combination of both

Step 3

Step 4:

From the SPSS output, we see our test statistic for the main effect of diet is F(2, 24) = 4.650 with a p-value = 0.020< .05. The test statistic for the main effect of exercise is F(1, 24) = 135.200 with a p-value = .000 < .05, and the test statistic for the interaction effect is F(2, 24) = 13.850 with a p-value of 0.000.

Step 5

Since the p–value for the main effect of exercise, diet and interaction between exercise and diet are all < 0.05, we fail to reject H_{0} and conclude that diet and exercise have significant effect on mental acuity. Specifically those who have 30%-60% fat diet has mean acuity (M=4.8) than do those who have<30% fat diet (M=4.1) and those who have >60% fat diet (M=3.700). Moreover, those who exercise 60% or more have mean acuity (M=5.933) greater than others who exercise less than 60 minutes (M=2.467).

B

The Tukey HSD shows that since p=0.016<0.05, those who take 30%-60% fat diet significantly differs from others

C

The effect sizes are the following: exercise = 0.849, diet= 0.279, and interaction = .536. While diet shows small effects, exercise and the interaction show a large effect.