Question

The USDA-recommended daily number of calories for a female aged 21-25 is 2200. A clinical trial is conducted to assess the effectiveness of a mindful-eating intervention program on caloric intake among overweight women. Although sixty participants were randomized into two groups at baseline, 5 participants dropped out of the study prior to completion. Group 1 received a mindfulness-based intervention where they were taught how to pay close attention to their feelings and emotions before they ate in order to prevent excess caloric intake. Group 2 were simply taught how to monitor and record their daily caloric intake. Mean daily caloric intake was assessed for 8 weeks. The results are summarized in the table below:

Groups	Sample Size	Mean Daily Caloric Intake
Mindfulness-based Caloric Intake Monitoring	30	1520.37 (146.8)
Non-mindfulness-Based Caloric Intake Monitoring	25	2069.54 (158.7)

According to this data, is there evidence at the 0.05 level of a statistically significant difference in the mean daily caloric intake for the mindfulness-based group versus the non-mindfulness-based group?

c. Indicate the appropriate test statistic

e. Compute and indicate the ratio of sample variances

f. Compute and indicate the pooled estimate of common standard deviation (Sp)

Answer 1

Claim : To test whether there is statistically significant difference in the mean daily caloric intake for the mindfulness-based group versus the non-mindfulness-based group

Hypothesis :

  

Two tailed test

Given that

For Mindfulness-based Caloric Intake Monitoring

n1 = 30 , S1 = 146.8

n2 = 25    S2 = 158.7

Before proceed to find the Confidence interval we have to check the population SD are equal or not

Hypothesis :

  

Two tailed test

Test statistic : F = ( larger sample variance / smaller sample variance )

  

DF for Numerator = n1-1 = 29

DF for denominator = n2-1 = 24

Critical value: Fc = F( 0.05,29,24) = 1.94....( From F table )

Decision Rule : We Fail to Reject Ho

Conclusion : There is sufficient evidence to conclude that the population variances are equal

Test Statistics :

   where Sp is pooled standard deviation

we get pooled estimate of common standard deviation Sp = 152.39 ........( ANSWER)