Jeralyn Jones
April 15, 2018
Research question
Does expression of affection on social media make couples feel happier in their relationship?
Data collection
Couples filled out a questionnaire on Facebook to the users who have the relationship status as "In a relationship”, “Engaged to” or “Married to”. The technique of simple random sampling is used to select a random sample of 30 participants.
Data
The data obtained regarding the happiness level and affection is given below:
Happiness_Don’t flaunt / Happiness_Flaunt-3 / -2
-3 / -3
-3 / -3
0 / 1
3 / -3
3 / -3
3 / -3
2 / 3
-3 / 1
3 / 3
2
3
1
-3
3
-3
-3
0
3
-3
Hypothesis testing
Assumptions
1)For the application of parametric test it is required that the sample is normally distributed.
From the histogram given above, it is evident that it is approximately bell-shaped indicating that the distribution is normal. Hence the assumption of normal distribution for application of the parametric test is satisfied.
2)With the help of Levene test it is important to check the assumption of equality of variance.
Test and CI for Two Variances: Happiness_Don’t flaunt, Happiness_Flaunt
Method
Null hypothesis σ(Happiness_Don’t flaunt) / σ(Happiness_Flaunt) = 1
Alternative hypothesis σ(Happiness_Don’t flaunt) / σ(Happiness_Flaunt) ≠ 1
Significance level α = 0.05
Statistics
95% CI for
Variable N StDev Variance StDevs
Happiness_Don’t flaunt 10 2.898 8.400 (2.402, 4.349)
Happiness_Flaunt 20 2.645 6.997 (2.328, 3.332)
Ratio of standard deviations = 1.096
Ratio of variances = 1.200
95% Confidence Intervals
CI for
CI for StDev Variance
Method Ratio Ratio
Bonett (0.851, 1.640) (0.725, 2.691)
Levene (0.754, 1.530) (0.569, 2.342)
Tests
Test
Method DF1 DF2 Statistic P-Value
Bonett — — — 0.470
Levene 1 28 0.13 0.725
Null Hypothesis, ho: there is no significant difference in the variance of two groups. The alternative hypothesis, h1: there is a significant difference in the variance of two groups. With (F=0.13, p>5
%), I fail to reject the null hypothesis and conclude thatthere is no significant difference in the variance of two groups.
Hypothesis
Step 1: hypothesis
The null hypothesis, Ho: there is no significant difference in the mean happiness level between those who don't flaunt and those who flaunt their love on social media.
The alternative hypothesis, h1: the mean happiness level for those who flaunt their love and social media is greater than those who don't flaunt.
Step 2: level of significance
The level of significance Alpha is taken as 5%.This is an indication that I am 95% confident about my results.
Step 3:test statistic
I use T-test for independent samples with equal variances to test my hypothesis.
Two-Sample T-Test and CI: Happiness_Don’t flaunt, Happiness_Flaunt
Two-sample T for Happiness_Don’t flaunt vs Happiness_Flaunt
N Mean StDev SE Mean
Happiness_Don’t flaunt 10 0.20 2.90 0.92
Happiness_Flaunt 20 -0.45 2.65 0.59
Difference = μ (Happiness_Don’t flaunt) - μ (Happiness_Flaunt)
Estimate for difference: 0.65
95% upper bound for difference: 2.45
T-Test of difference = 0 (vs <): T-Value = 0.61 P-Value = 0.728 DF = 28
Both use Pooled StDev = 2.7291
T= 0.61
Step 4: P-value
P value is the probability that null hypothesis is true.P-Value = 0.728
Step 5: Decision rule
The decision rule is to reject the null hypothesis if P-value is less than Alpha, 5%. Else if P-value is greater than 5%, I fail to reject the null hypothesis.
Step 6: conclusion
Since P value is greater than 5%, I fail to reject the null hypothesis and conclude thatthere is no significant difference in the mean happiness level between those who don't flaunt and those who flaunt their love on social media.
Conclusion
With (t=0.61, p>5%), I can say that there is no significant difference in the mean happiness level between those who don't flaunt and those who flaunt their love on social media.