SPSS Individual Coursework

Question 1

(a)
1.1 Group Statistics

method
N
Mean
Std. Deviation
Std. Error Mean output 1 method1
10
11.860
4.7782
1.5110

2 method2
10
13.410
4.3309
1.3695

Checking assumptions:
Two sample t-test assumes equal variances. Check with Levene’s test
H0: 12 = 22 H1: 12  22
Presented in the output above, the significance level for Levene’s is 0.807. This means that no evidence against equal variances, therefore, use the one provided in the first line of the table, which refers to equal variances assumed.

Assessing differences between the groups:
Main test
H0: 1 = 2 H1: 1  2 n1, n2<30
95% CI for difference 1 – 2 is -5.83 to 2.73, includes 0 but goes down to -5.83 output (measure 2 could make almost 5.83 output more)
The significance (two-tailed) value is 0.457; no evidence against null hypothesis. There is not a statistically significant difference in the mean output for method one and two.

The results of the analysis:
An independent samples t-test was conducted to compare the average output for method one and method two. There was no significant difference in output for method one (mean=11.860, std deviation=4.7782) and method two (mean=13.410, std deviation=4.3309; t (18) =-0.760, p value=0.457 (2-tailed). The magnitude of the differences in the means (mean difference=1.55, 95% confidence interval:-5.83 to 2.73)
(b)

2.1 Paired Samples Statistics

Mean
N
Std. Deviation
Std. Error Mean
Pair 1 method1 11.860
10
4.7782
1.5110

method2
13.410
10
4.3309
1.3695

2.2 Paired Samples Correlations

N
Correlation
Sig.
Pair 1 method1 & method2
10
-.140
.699

2.3 Paired Samples Test

Paired Differences t df
Sig. (2-tailed)

Mean
Std. Deviation
Std. Error Mean
95% Confidence Interval of the Difference

Lower
Upper

Pair 1 method1 - method2
-1.5500
6.8842
2.1770
-6.4746
3.3746
-.712
9
.495

Correlation
Correlation indicated efficacy of paring, here is -0.14, which is very small and negative paired correlation, so it suggests that paired test is not appropriate to analyze the data, the independent sample size would be more appropriate to analyze.

Determining overall significance
Given in the output above, the p value is 0.495, so there is no significant difference between the two methods.

Comparing mean values
The mean output by using method 1 was 11.86 and the mean output by using method 2 was 13.41; the mean difference in the two methods was 1.55 with a 95% confidence interval ranging from -6.47 to 3.37, includes 0 but goes down to -6.47, measure 2 could make almost 6.47 more output. It seems that there is a significant difference between the two methods. Such results may occur since the sample size is not enough big. Therefore, it suggests to take a bigger random sample and to get a further investigation.
Moreover, there are many other factors that may have also influenced the increase in output. For example, the skills owned by different operatives could have contributed.

The results of the analysis:
A paired sample t-test was conducted to evaluate the impact of the two methods on output. According to the p value, it seems that no significant difference between the two methods, while considering the means difference along with the big range of 95% confidence interval, it seems that there is some significant difference. So it suggests that to take a bigger sample size to get further investigate.
(c)

Here, we use Mann Whitney U test and Median test as non-parametric method for two sample tests.
The reason to choose these methods to analyze the data:
Correlation indicated efficacy of paring, here is -0.14, which is very small and negative paired correlation, so it suggests that paired test is not appropriate to analyze the data, the independent sample size would be more appropriate to analyze.

The Mann Whitney U test is designed to test for differences between two independent methods on a continuous measure. Instead of comparing means of the two methods, it actually