Statistical Analysis: T-test Worksheet for Diabetes Intervention

This assignment analyzes the effect of a diabetes education intervention on the number of diabetes diagnoses in pre-diabetic children using t-tests. The assignment includes two parts: an independent samples t-test comparing a diabetes education group to a control group, and a dependent samples t-test examining changes in the diabetes education group over two years. The student calculates t-statistics, determines critical values, and draws conclusions based on hypothesis testing, including null and alternative hypotheses. The solution includes calculations of means, variances, and degrees of freedom, demonstrating the application of t-tests to real-world data and interpretation of statistical results.

Running header: T-test 1
T-tests
Name:
Institution:

T-test 2
INDEPENDENT SAMPLES: An exercise physiologist is interested in examining the
effect of a diabetes education intervention (designed to teach parents about healthy eating and
physical activity) on the number of diabetes diagnoses that occur within the following year in a
sample of pre-diabetic children. A total of 12 families with pre-diabetic children are recruited.
Six families were in the diabetes education group and 6 were in the control group. The number
of diagnoses for each family are recorded in the Table below.
Family Group
# of diagnoses
in year 1
1 Diabetes education 10
2 Diabetes education 8
3 Diabetes education 5
4 Diabetes education 4
5 Diabetes education 6
6 Diabetes education 10
7 Control 7
8 Control 8
9 Control 9
10 Control 6
11 Control 5
12 Control 9
1. What is your null hypothesis for this research question?
Null hypothesis: There is no mean difference in the number of diabetes diagnoses
between the diabetes education intervention and the control groups.
Alternative hypothesis: There is a mean difference in the number of diabetes diagnoses
between the diabetes education intervention and the control groups.
H0 :μ1 =μ2
H1 : μ1 ≠ μ2
2. Using an independent samples t-test, calculate the observed t-statistic from the
data above in Excel. Hint: You’ll need to calculate the mean and sum of squares for each group.

T-test 3
Mean for the diabetes education intervention group
μ1= ∑ xi
N = 10+8+5+ 4+ 6+10
6 =7.167
Mean for Control group
μ2= ∑ xi
N = 7+8+ 9+6+5+ 9
6 =7.333
Variance for Diabetes education intervention
diabetes education intervention x-μ (x-μ)^2
10 2.833 8.027777778
8 0.833 0.694444444
5 -2.167 4.694444444
4 -3.167 10.02777778
6 -1.167 1.361111111
10 2.833 8.027777778
Total 32.833
σ 1
2=∑ ( x −μ )2
n−1 = 32.833
5 =6.5667
Variance for Control group
diabetes education intervention x-μ (x-μ)^2
10 -0.333 0.111111
8 0.667 0.444444
5 1.667 2.777778
4 -1.333 1.777778
6 -2.333 5.444444
10 1.667 2.777778
Total 13.333
σ 2
2=∑ ( x −μ )2
n−1 = 13.333
5 =2.6667
Test statistics
t= μ1−μ2
√ σ 1
2
n1
+ σ2
2
n2
= 7.167−7.333
√ 6.5667
6 + 2.6667
6
=−0.166
1.241 =−0.1344

T-test 4
3. What is the critical t-value for this research question? Hint: Use the table of
critical values and the degrees of freedom to locate this value. The alpha = 0.05.
Degrees of freedom = n-2 = 12-2 = 10
t0.05,10=2.22 81
4. What can you conclude based on your data? Do you reject or fail to reject the
null hypothesis?
Decision rule: If the t-calculated is less than the critical value we fail to reject the null
hypothesis.
Conclusion: It is evident that the t-calculated -0.1344 is less than the critical value 2.2281
thus we fail to reject the null hypothesis and conclude that there is no mean difference in
the number of diabetes diagnoses between the diabetes education intervention and the
control groups

T-test 5
DEPENDENT SAMPLES: Using the same groups as above, the researcher continues
the intervention for a second year and wants to know if the number of diagnoses varies between
year 1 and year 2. Only the 6 families receiving the intervention are followed. The number of
diagnoses for each family are recorded in the Table below.
Family Group
# of diagnoses in
year 1
# of diagnoses in
year 2
1 Diabetes education 10 6
2 Diabetes education 8 5
3 Diabetes education 5 6
4 Diabetes education 4 4
5 Diabetes education 6 2
6 Diabetes education 10 3
1. What is your null hypothesis for this research question?
Null hypothesis: There is no mean difference in the number of diabetes diagnoses for the
diabetes education intervention between year 1 and year 2.
Alternative hypothesis: There is a mean difference in the number of diabetes diagnoses
for the diabetes education intervention between year 1 and year 2.
H0 :μ=0
H1 : μ ≠ 0
2. Using an dependent samples t-test, calculate the observed t-statistic from the data above
in Excel. Hint: You’ll need to calculate the sum of differences and the sum of squared
differences for each group.
Family year 1 year 2 Difference (x) (x-μ)^2
1 10 6 4 1.166667 1.361111
2 8 5 3 0.166667 0.027778
3 5 6 -1 -3.83333 14.69444
4 4 4 0 -2.83333 8.027778
5 6 2 4 1.166667 1.361111
6 10 3 7 4.166667 17.36111
Total 17 42.833

T-test 6
Mean for the difference
μ= ∑ xi
N = 4 +3−1+0+4 +7
6 =2.833
Standard deviation of difference =
σ = √ ( x−μ )2
n−1 = √ 42.833
5 =2.927
Test statistics
t= μ1 −0
σ
√n
= 2.833−0
2.927
√6
=2.3712
3. What is the critical t-value for this research question? Hint: Use the table of critical
values and the degrees of freedom to locate this value. The alpha = 0.05.
Degrees of freedom = n−1=6−1=5
t0.05 ,5=2. 5706
4. What can you conclude based on your data? Do you reject or fail to reject the null
hypothesis?
Decision rule: If the t-calculated is less than the critical value we fail to reject the null
hypothesis.
Conclusion: It is evident that the t-calculated 2.3712 is less than the critical value 2.5706
thus we fail to reject the null hypothesis and conclude that there is no mean difference in
the number of diabetes diagnoses for the diabetes education intervention between year 1
and year 2.