Anova Chi Square Test Homework

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
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
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 R2 is 0.011. this means that 1.1% of variation in person’s life satisfaction is explained by hours in a week spent exercising

ANOVAa
Model Sum of Squares df Mean Square F Sig.
1 Regression 1.252 1 1.252 .195 .664b
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 u1=mean of people taking pharmaceuticals and u2 is the mean of those taking the new relaxation technique, we have:

H0­­: u1-u2=0

H1­­: u1-u2<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 H0. 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 u1 be the mean of life satisfaction rating for respondents who have no interaction with their mother, u2 =the mean of life satisfaction rating for respondents who have low interaction with their mother u3= the mean of life satisfaction rating for respondents who have moderate interaction with their mother and u4= the mean of life satisfaction rating for respondents who have high interaction with their mother, we have:

H0:u1=u2=u3=u4

H1: 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 H0. 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 Ho. 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

H0: mental acquity do not vary by diet, exercise or a combination of both

Ha:  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 H0 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.