Free Fall Quarter: 2014 Essay Example

Independent-Samples t-Test with SPSS

Independent-Samples t-Test with SPSS
In this paper we will describe the basics of hypothesis testing using SPSS. For the purposes of this assignment, the two chosen variables were SEX and AGE. The proposed hypothesis is: On this sample, men and women have a different mean age.

When comparing mean age, the appropriate statistic is Student’s t. The assumptions for running a t-test are (Cressie & Whitford, 1986):

The data are continuous.
The data follows the normality probability distribution.
For a two-sample independent t-test with unequal variances, the individual variances on the two compared groups are statistically equal.
The two groups were sampled independently: there is no relationship between the individuals from a sample when compared with the other.

A random sampling method was used to collect the data.

Null hypothesis: There is no significant difference in age between males and females.
H0: μ1=μ2

Alternative hypothesis: There is a significant difference in age between males and females.Ha: μ1≠μ2

Where μ1 = mean age of males; and μ2 = mean age of women.
Set level of significance alpha: α=0.05

Perform independent-samples t-test using SPSS:

The histogram shows a fairly normally distributed sample. It is a bit right skewed, but the Kolmogorov-Smirnov test provides a statistically significant p-value of .000 (less than .0001), which means that our sample population and a hypothetical normal distribution are not statistically different. Therefore, we comply with the normality assumption.

Next, SPSS reports the output for the t-test:

According to this sample, men are on average 45.87 years old, while women are 46.54 years old.
SPSS performs a Levene’s Test for Equality of Variances, which results non-significant (p-value 0.87). This means that we assume that both groups were sampled from the same populations and thus their variances are statistically equal, which was another of the t-test assumptions.
Last, the t-test reports a test statistic of -0.799 (p-value 0.444), which means that on average and at a 5% significance level, the age between men and women are not statistically different. Based on this sample, we fail to reject the null hypothesis.

Furthermore, to explore effect size, Cohen’s d was calculated (Thalheimer & Cook, 2002):

mean1: 45.87, sd1: 16.204, var1: 262.569616,
mean2: 46.54, sd2: 17.095, var2: 292.239025,
pooled sd: sqrt((variance1 + variance2)/2)
pooled sd: sqrt((262.569616 + 292.239025)/2)
pooled sd: 16.66
d = (mean1 - mean2)/pooled sd
d = (46.54 - 45.81) / 16.66
d = 0.043
A Cohen’s d of 0.043 can be in this case interpreted as the difference between the means of the groups standardized by the pooled standard deviation. With means around 45-46, and a mean difference of only 0.043, we can assume that the effect size is very small. In fact, SPSS reports the 95%CI of the difference as not significant, because it crosses the null value of zero (95%CI of the difference -2.358 , 1.033).

References

Cressie, N. A. C., & Whitford, H. J. (1986). How to Use the Two Sample t‐Test.Biometrical Journal, 28(2), 131-148.
Pallant, J. (2010). SPSS survival manual: A step by step guide to data analysis using SPSS. McGraw-Hill International. pp. 54, 63.
Thalheimer, W., & Cook, S. (2002). How to calculate effect sizes from published research articles: A simplified methodology. Retrieved January 23, 2015 from http://work-learning.com/effect_sizes.htm.