Question

Independent -Samples t-test (Worth 15 points) Scenario: An article in the journal Applied Nutritional Investigation reported the results of a comparison of two different weight-loss programs (Liao, 2007). In the study, obese participants were randomly assigned to one of two groups: (1) the soy group, a low-calorie group that ate only soy-based proteins or (2) the traditional group, a low-calorie group that received 2/3 of their protein from animal products and 1/3 from plant products. One of the dependent measures collected was the amount of body fat loss as a percentage of initial body weight. Fictional data on the percent of body fat loss appear in the table. Table: Diets Traditional Diet Soy Diet 2.5 1.21 0.86 1.7 1.78 2.5 1.23 1.9 0.50 2.0 1.76 3.2 (a) Calculate a 95% confidence interval and assess the research hypotheses. (8 pts) (b) Compute the effect size for this study and interpret its meaning. (7 pts)

Answer 1

Probability Plot of Soy Diet, Traditional Diet Normal 95% CI -1 1 2 Traditional Diet Soy Diet Soy Diet 99 99 Mean 2.3 StDev 0.5477 N 6 95 95 AD 0.300 90 90 P-Value 0.456 Traditional Diet 80 80 Mean 1.223 70 70 StDev 0.5007 60 60 N 6 50 50 AD 0.267 40 40 P-Value 0.538 30 30 20 20 10 10 1 1 1 2 4 Percent n

From above probability plots, it is clear that the two sets of observations come from two independent normal populations.

Test for Two Variances: Soy Diet, Traditional Diet

Method

Null hypothesis         Sigma(Soy Diet) / Sigma(Traditional Diet) = 1
Alternative hypothesis Sigma(Soy Diet) / Sigma(Traditional Diet) not = 1
Significance level      Alpha = 0.05

Statistics

Variable            N StDev Variance
Soy Diet         6 0.548     0.300
Traditional Diet    6 0.501     0.251

Ratio of standard deviations = 1.094
Ratio of variances = 1.197

                                                  Test
Method                          DF1 DF2 Statistic P-Value
F Test (normal)                   5    5       1.20       0.849

Since P-value>0.05 so we can assume that two population variances are same.

$95\%~CI~for~difference~between~two~diets:\\\\ \left(\bar{x}_1-\bar{x}_2-t_{0.025,10}\times s_p\times \sqrt{\frac{2}{6}}, ~\bar{x}_1-\bar{x}_2+t_{0.025,10}\times s_p\times \sqrt{\frac{2}{6}}\right )\\\\ =(0.402, 1.752)\\\\ where,~\bar{x}_1=sample~mean~for~Soy~diet=2.300\\\\ \bar{x}_2=sample~mean~for~Traditional~diet=1.223\\\\ s_1=sample~sd~for~Soy~diet=0.548\\\\ s_2=sample~sd~for~Traditional~diet=0.501\\\\ s_p=\sqrt{\frac{(6-1)s_1^2+(6-1)s_2^2}{10}}= 0.5247\\\\ Since~the~confidence~interval~contains~only~positive~values~so~we~are~95\%~confident~that\\ amount~of~body~fat~loss~due~to~Soy~diet~is~significantly~greater~than~amount~of~body~\\ fat~loss~due~to~Traditional~diet.\\\\ (b)~Effect~size=\frac{\bar{x}_1-\bar{x}_2}{s_p}=2.05~hence~it~is~large~effect~size.\\\\$

Problem 2 1. Population Students of math class. 2. F - distribution 3. i. The experimental errors of the observations are normally distributed i. Equal variances between treatments ii. Each sample is randomly sel ected and independent. 4. Assumption (iii) 5.

Probability Plot of Test score Normal 95% CI 99 Мean 81.33 StDev 9.619 N 15 95 0.332 AD 90 P-Value 0.469 80 70 60 50 40 30 20 10 5 1 50 70 80 90 100 110 120 Test score Percent 60

From above plot, we see that assumption of normality holds.

Test for Equal Variances: Test score versus Group

95% Bonferroni confidence intervals for standard deviations

    Group N    Lower    StDev    Upper
Re-write 5 3.94553 7.30068 28.0996
Self-quiz 5 3.48988 6.45755 24.8544
    Teach 5 4.61114 8.53229 32.8399

Bartlett's Test (Normal Distribution)
Test statistic = 0.28, p-value = 0.868

Levene's Test (Any Continuous Distribution)
Test statistic = 0.03, p-value = 0.966

Since p-value>0.05 so assumption of equal variance holds.

$6.~Let~\mu_t=true~mean~score~of~Group~"Teach"\\\\ \mu_r=true~mean~score~due~of~"Re-write"\\\\ \mu_s=true~mean~score~due~of~"Self-quiz"\\\\ Null~hypothesis,~H_0:\mu_t=\mu_r=\mu_s\\\\ 7.~Alternative~hypothesis,~H_A:~At~least~one~\mu~is~different~from~others.\\\\ 8.~Between~df=3-1=2,~With~df=15-3=12\\\\ 9.~Critical~cut-off~point=F_{0.05,2,12}=3.8853\\\\ 10.~\bar{x}_{10}=mean~of~Group~"teach"=81.4;~ \bar{x}_{20}=mean~of~Group~"Re-write"=73.4;~ \bar{x}_{30}=mean~of~Group~"Sel-quiz"=89.2,~\bar{x}_{00}=grand~mean\\ =81.3333,~\sum_{i=1}^3\sum_{j=1}^5x_{ij}^2=100522~where,~x_{ij}=jth~observation~from~ith~group.\\\\$

$SST=sum~of~square~due~to~total=\sum_{i=1}^3\sum_{j=1}^5x_{ij}^2-15\bar{x}_{00}^2=1295.415\\\\ SSB=sum~square~between~groups=5\sum_{i=1}^3(\bar{x}_{i0}-\bar{x}_{00})^2=624.1333\\\\ MSB=mean~sum~square~between~groups=SSB/2=312.0666\\\\ SSW=sum~square~within~group=SST-SSB=1295.415-624.1333=671.2817\\\\ MSW=mean~sum~square~within~group=SSW/12=55.9401\\\\ F-ratio=\frac{MSB}{MSW}=5.5786\\\\ \begin{matrix} Source &SS & DF &MS &F~ratio \\ Between&624.1333 &2 & 312.0666 &5.5786 \\ Within &671.2817 &12 &55.9401 & \\ Total&1295.415 & 14 & & \end{matrix}\\\\ 11.~Since~F-ratio>Critical~cut-off~point~so~we~reject~the~null~hypothesis.$