Using Excel to Run a ttest on the context of a Case Study of Inferential Statistics
The purpose of this study is to assess if there is a significant difference between web based and personal training for an IT program, based on the information collected from a sample.
The data was collected from 25 students undertaking different classes, some of which were webbased and some other had personal training. An average for both webbased and personal training grades was obtained and recorded. Also, other variables such as Education Level, Experience and Gender were recorded
The original data set is shown in the table below:

Grade  Web 
Grade  Personal 
Education Level 
Experience 
Gender  
1 
60% 
78% 
1 
5 
1 
2 
61% 
82% 
2 
6 
1 
3 
64% 
83% 
2 
5 
2 
4 
65% 
79% 
2 
15 
2 
5 
69% 
78% 
1 
13 
1 
6 
70% 
78% 
3 
15 
2 
7 
72% 
73% 
1 
8 
1 
8 
72% 
77% 
2 
11 
1 
9 
74% 
75% 
2 
12 
2 
10 
78% 
76% 
2 
14 
1 
11 
78% 
100% 
3 
14 
1 
12 
77% 
77% 
3 
13 
1 
13 
79% 
78% 
3 
12 
2 
14 
82% 
77% 
1 
14 
2 
15 
86% 
79% 
3 
15 
2 
16 
85% 
100% 
3 
17 
1 
17 
89% 
89% 
2 
5 
1 
18 
95% 
87% 
4 
6 
2 
19 
94% 
88% 
3 
5 
2 
20 
95% 
90% 
3 
7 
2 
21 
93% 
87% 
3 
8 
1 
22 
97% 
100% 
4 
15 
2 
23 
97% 
88% 
4 
15 
1 
24 
96% 
87% 
3 
17 
2 
25 
99% 
84% 
4 
30 
1 
Education level 
Gender 

1  High School 
1  Male 

2  College 
2  Female 

3  University 

4  Post Graduate 

Let's perform some statistical analyses. We first show the descriptive statistics
Summary Measures for Web Training  
Mean 
0.8108 
Standard Error 
0.03 
Median 
0.79 
Mode 
0.72 
Standard Deviation 
0.13 
Sample Variance 
0.02 
Kurtosis 
1.30 
Skewness 
0.11 
Range 
39% 
Minimum 
60% 
Maximum 
99% 
Sum 
20.27 
Count 
25 
The mean score for Web Training is equal to
\[\bar{X}=\frac{1}{n}\sum{{{X}_{i}}}=\text{81}\text{.08 }\!\!%\!\!\text{ }\]and the standard deviation is equal to
\[s = \frac{1}{n1}\sum{{{\left( {{X}_{i}}\bar{X} \right)}^{2}}}=13%\]The data range from 60% to 99%.
· Now we show a summary for Personal Training
Summary Measures for Personal Training  
Mean 
0.836 
Standard Error 
0.016 
Median 
0.820 
Mode 
0.780 
Standard Deviation 
0.079 
Sample Variance 
0.006 
Kurtosis 
0.065 
Skewness 
0.912 
Range 
27% 
Minimum 
73.0% 
Maximum 
100% 
Sum 
20.9 
Count 
25 
The mean score for Personal Training is equal to
\[\bar{X}=\frac{1}{n}\sum{{{X}_{i}}}=\text{83}\text{.6 }\!\!%\!\!\text{ }\]and the standard deviation is equal to
\[s=\frac{1}{n1}\sum{{{\left( {{X}_{i}}\bar{X} \right)}^{2}}}=7.9%\]The data range from 73% to 100%.
Correlation Analysis: Let's perform a correlation analysis between the variables Webbased training grade, and Personal Training grade. For this purpose, we display an output from Statdisk:
We observe that the correlation coefficient is equal to
\[R=0.54466\]Since the critical values for \(R\) are \(\pm 0.39607\), for a sample size of \(n=25\), we conclude that the correlation is significantly different from zero (positive). Also, we find that the coefficient of determination is equal to
\[{{R}^{2}}=0.29665\]which implies that approximately 29.665% of the variation in Personal Training grade is explained by Webbased training grade.
Histograms: The following histograms correspond to Web and Personal Training Scores
There shape of the histograms is mildly normal (bellshaped), but we need further analysis to determine whether variables are normally distributed or not.
To this end, we use Minitab to perform a normality test. The results are shown in the diagram shown below
The pvalue for test is 0.082, which is less than the significance 0.05. This means that we fail to reject the null hypothesis of normality.
Confidence Interval: We have that the 95% confidence interval for \({{\mu }_{D}}\) is given by
\[ CI=\left( \bar{D}{{t}_{c}}\times \frac{{{s}_{D}}}{\sqrt{n1}},\text{ }\bar{D}+{{t}_{c}}\times \frac{{{s}_{D}}}{\sqrt{n1}} \right) = \left( 32.064\times \frac{10.57}{\sqrt{24}},\text{ }3+2.064\times \frac{10.57}{\sqrt{24}} \right) \] \[ =\left( 1.45327,\text{ }7.45327 \right) \]This means that there's a 95% probability that this interval contains the actual population mean difference \({{\mu }_{D}}\).
Inferential Statistics: Our purpose is to determine if there is a significant difference between web based and personal training. Since we have paired samples our objective is twofold:
 Use hypothesis testing to determine whether the mean difference between the scores is significantly different from zero.
 Use the sample data to construct a confidence interval to estimate \({{\mu }_{D}}\), the population mean difference \({{\mu }_{D}}\) of the scores.
First, we need the table with the difference between the scores
Grade  Web 
Grade  Personal 
Difference$(D)$  
1 
60% 
78% 
18% 
2 
61% 
82% 
21% 
3 
64% 
83% 
19% 
4 
65% 
79% 
14% 
5 
69% 
78% 
9% 
6 
70% 
78% 
8% 
7 
72% 
73% 
1% 
8 
72% 
77% 
5% 
9 
74% 
75% 
1% 
10 
78% 
76% 
2% 
11 
78% 
100% 
22% 
12 
77% 
77% 
0% 
13 
79% 
78% 
1% 
14 
82% 
77% 
5% 
15 
86% 
79% 
7% 
16 
85% 
100% 
15% 
17 
89% 
89% 
0% 
18 
95% 
87% 
8% 
19 
94% 
88% 
6% 
20 
95% 
90% 
5% 
21 
93% 
87% 
6% 
22 
97% 
100% 
3% 
23 
97% 
88% 
9% 
24 
96% 
87% 
9% 
25 
99% 
84% 
15% 
Average \((\bar{D})\) 
3%  
St. Dev \(({{s}_{D}})\) 
10.57% 
We need to perform a paired ttest to determine whether the difference is statistically significant or not. We want to test the following hypotheses:
\[\begin{array}{cc} & {{H}_{0}}:{{\mu }_{D}}=0 \\ & {{H}_{A}}:{{\mu }_{D}}\ne 0 \\ \end{array}\] Assumptions: In order to apply the ttest we need to show that the differences are normally distributed. We tested for normality at the 0.05 significance level, and we found that we cannot reject the null hypothesis of normality at the 0.05 significance level.
We compute the ttest statistics as
\[t = \frac{\bar{D}{{\mu }_{D}}}{{{s}_{D}}/\sqrt{n1}}=\frac{30}{10.57/\sqrt{24}} = 1.390439\]The critical value for the twotailed tdistribution, with 24 degrees of freedom, for \(\alpha =0.05\), is equal to
\[{{t}_{c}}=2.064\]Remarks:
 The sample size is not big enough to approximate by a normal distribution. It would be recommended to increase the sample size.
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