The usual treatment of hypothyroidism involves daily thyroid
The usual treatment of hypothyroidism involves daily thyroid replacement with levothyroxine (L-T4). Optimal therapy is monitored by measurement of serum TSH. The normal thyroid gland secretes both levothyroxine (L-T4) and levotriiodothyronine (L-T3). It is generally believed that treatment L-T4 alone results in normal levels of serum L-T3 to achieve a euthyroid state. Some studies suggest, however, that patients feel better when taking an L-T4/L-T3 combination compared with L-T4 alone.
Woeber (2002) wished to study this clinical issue further by evaluating the effect of L-T4 replacement therapy on serum free T4. He studied the relationship in a historical cohort of patients with chronic autoimmune thyroiditis and a group of normal individuals. Woeber\'s data gave the following results.
At the 1% level, is there a difference in the mean free T4 equal among the three groups of subjects. Why or why not?
Is there a difference in the average ? Explain. If needed, you may use a 95% level of confidence.
You must include in your work:
1. Test statistic, critical value or p-value, decision, and interpretation.
2. Confidence interval end points and decision.
| SUMMARY | |||||
| Groups | Count | Sum | Average | Variance | |
| A: Control | 20 | 271 | 13.55 | 2.95 | |
| B: Normal Patients | 18 | 252 | 14 | 4.235 | |
| C: Patients on Therapy | 35 | 556 | 15.886 | 3.987 | |
| ANOVA | |||||
| Source of Variation | SS | df | MS | F | P-value |
| Between Groups | 0.00003487 | ||||
| Within Groups | 246.493 | ||||
| Total | 330.493 | ||||
Solution
Complete the anova table as follows:
As p value is 0.05, reject null hypothesis.
The means of all groups are not equal.
| Task | Task time per worker | Number of workers | ||||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | |
| Between Groups | 84 | 2 | 42 | 13.971999 | 3.49E-05 | |
| Within Groups | 246.493 | 82 | 3.0060122 | |||
| Total | 330.493 | 84 | ||||
