Question

Problem 3: A study examines chocolate's effects on blood vessel function in healthy people. In the randomized, double-blind, placebo- controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of 2.32, while the control group had a mean change of -0.96 with a standard deviation of 1.58. a. Find a 95% confidence interval for the difference in means between the two groups, where he represents the mean increase in flow-mediated dilation for people eating dark chocolate every day and represents the mean increase in flow- mediated dilation for people eating a dark chocolate substitute each day. You may assume that neither sample shows significant departures from normality. Please show your work including the formula that you are using. Round your answers to two decimal places. (5 points) Note: For a 95% confidence interval, t* = 2.262. b. Provide an appropriate interpretation in the context of this problem) of your 95% confidence interval. (3 points)
show work please

Answer 1

a)

Sample #1   ---->   1              
mean of sample 1,    x̅1=   1.300              
standard deviation of sample 1,   s1 =    2.320              
size of sample 1,    n1=   11              
                      
Sample #2   ---->   2              
mean of sample 2,    x̅2=   -0.960              
standard deviation of sample 2,   s2 =    1.580              
size of sample 2,    n2=   10              
                      
difference in sample means =    x̅1-x̅2 =    1.3000   -   -1.0   =   2.26

DF = min(n1-1 , n2-1 )=   9              
t-critical value , t* =    2.2622   (excel formula =t.inv(α/2,df)          
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.860          
margin of error, E = t*SE =    2.2622   *   0.860   =   1.94
                  
difference of means = x̅1-x̅2 =    1.3000   -   -0.960   =   2.260

confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    2.2600   -   1.9446   =   0.3154
Interval Upper Limit = (x̅1-x̅2) + E =    2.2600   -   1.9446   =   4.2046

CI ( 0.32 , 4.20)

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b)

one can be 95% confident that true mean difference between groups will lie between these limits

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