Researchers were interested in comparing the long-term psychological effects of being on a high-carbohydrate, low-fat(LF) diet versus a high-fat, low-carbohydrate (LC) diet. A total of 106 over-weight and obese participants were randomly assigned to one of these energy-restricted diets. At 52 weeks, 32 LC dieters and 33 LF dieters remained. Mood was assessed using a total mood disturbance score (TMDS), where a lower score is associated with a less negative mood. A summary of these results follows:

This is the data table: http://prntscr.com/fynm9f

Is there a difference in the TMDS at week 52? Test the null hypothesis that the dieters’ average mood in the two groups is the same. Use significance level of 5%?

Critics of this study focus on the specific LC diet (that is, the science) and the dropout rate. Explain why the dropout rate is important to consider when drawing conclusions from this study.

This is the final question of my homework assignment and I am completely lost. Please help, state your work and formulas. Thank you. For part a, I have no idea. But for part b, drop out rate is important because 1. It may provide useful information. and 2.If dropout rate is too large, we have to use t-test since the remaining sample is too small. Otherwise we may use z-test.

Assuming that data are nearly normally distributed, you can use a Welch separate variances t test to test $H_0: \mu_1 = \mu_2$ against $H_0: \mu_1 \ne \mu_2,$ where $\mu_1$ and $\mu_2$ are population means for subjects on the LF and LC diets.

Strictly speaking, because population variances are unknown, this should be a t test instead of a z test, regardless of sample sizes. (However, a z test might be a reasonable approximate test, for large enough sample sizes.) Because you do not know that the population variances are equal, this should be a Welch 2-sample t test, not a pooled 2-sample t test.

I will leave it for you to find and use the formulas for the Welch test. Results from Minitab 17 are shown below, so that you can verify your results.

Sample   N  Mean  StDev  SE Mean
1       32  47.3   28.3      5.0
2       33  19.3   25.8      4.5

Difference = μ (1) - μ (2)
Estimate for difference:  28.00
T-Test of difference = 0 (vs ≠):
T-Value = 4.16  P-Value = 0.000  DF = 62


The P-value $< .0.5$ indicates you should reject $H_0.$ [For a pooled 2-sample t test, Minitab gives $T = 4.17$, a P-Value < 0.0005, 63 degrees of freedom and a pooled SD $S_p = 27.059.$] Clearly, LC patients who stayed in the study are unhappy about something, perhaps the diet.

However, the drop-out rate of about 40% of the original 106 subjects may indicate a bad mood for subjects on both diets. No diet is likely to be successful in practice if people won't adhere to it.

I don't know the scale on the TMDS, but I would want to have a look at the data to see if they are approximately normal. Especially low scores, such as in the LC group, may be skewed to the right and thus not normal. But the difference between average scores of LC and LF subjects is so large that lack of normality (even if present) may not be a serious issue.

Note: You will get a cheerier reception on this site if you show some work of your own and avoid giving orders such as 'state your work and formulas'. We are here to help you understand mathematics and statistics, not to do you homework for you.

• I was not given a bunch of data - just the table with variables already calculated. – satjav Jul 21 '17 at 21:34
• Is there a way to do this problem without the Welch Test? I have looked it up and it is not something that we have learned yet. – satjav Jul 21 '17 at 21:36
• In my view there is no correct way to analyze these data without using a Welch test. A pooled two-sample t test may be a reasonable approximation because ($n_1 \approx n_2$). And the difference in means is so large that the conclusion happens to be the same. I gave results for a pooled test. See if you get the same results from formulas in your book. // Along with giving you 'a bunch of data', did your book say anything about assuming the data to be nearly normal? If not, I give he authors about a C-. – BruceET Jul 21 '17 at 21:47