Independent-Samples T-Test Computations please help?
You obtain the data shown below for 30 girls with anorexia who were treated with psychodynamic therapy
Data:
0.3 1.7 –1.2 2.2 –1.8 0.9 2.6 –3.2 0 –0.5
–4.2 –8.6 0.1 0.6 8.2 0 –2.8 –5.0 0.8 5.0
5.8 0.3 0.6 –5.9 1.2 –7.8 –9.2 1.3 1.9 0.4
Report the appropriate null and research hypotheses and the critical value
(alpha = .05). In addition, provide your statistical conclusion, including calculation of test statistic.
Please help me if you can? My book just does not help me
Posted in psychodynamic psychotherapy
content rss
February 21st, 2010 at 10:57 am
The null hypothesis is generally given as some version of "no change." For this problem the null hypothesis would be the treatment had no effect on the girls. Numerically, it may be mu = 0, if 0 is the "current condition" value.
The null hypothesis is generally given as some sort of change, sometimes change in either direction (a two-tailed test), sometimes a change in one direction (a one-tailed test). For this problem, it seems the concern is whether the treatment helped the anorexic girls, so it would be a one-tailed test (that is, if the treatment had a negative effect we will reject it, just as if it had no effect). If positive values are improvement, the alternate hypothesis would be mu > 0.
For the sample size given, 30, you could use a normal approximation, but since you specify the t-test, the critical value is 1.699 (one-tailed test; alpha = 0.05; n = 30, so n-1 or 29 degrees of freedom), assuming positive values indicate improvement. The critical value would be -1.699 if improvement were noted by negative values.
We now need to find the mean of the data set and determine if that mean is statistically significant, that is sufficiently far from a baseline to say the treatment offered improvement. The mean is -0.5433.
If positive scores indicate improvement, there is no need to go further with this problem, as the treatment appears to have had a slight negative effect on the patients.
Changing thinking to assume 0 is the norm and negative scores indicate improvement, we need to check if there is enough improvement to say it is caused by the treatment and not some random coincidence or other factor. To do this we must calculate a t-score and compare it to the critical value. A t-score farther away from 0 than the critical value is significant at the 5% level and would indicate the treatment is beneficial.
t = (x-bar - mu)/(s/sqrt(n))
t = (-0.5433 - 0)/(3.9856/5.4772)
t = -0.75
Because this value is closer to zero than the critical value, it appears the treatment has no statistically significant effect.