Twelve dieters lost an average of 13.7 pounds in 6 weeks when given a special diet plus a "fat-blocking" herbal formula. A control group of twelve other dieters were given the same diet, but without the herbal formula, and lost an average of 10.7 pounds during the same time. The standard deviation of the "fat-blocker" sample was 2.6 and the standard deviation of the control group was 2.4. Find the 95% confidence interval for the differences of the means.

The 95% confidence interval is  

Step-by-step explanation:

From the question we are told that

   The  sample mean for fat-blocking

     The  sample size  for fat-blocking  

      The standard deviation for  fat-blocking is

      The  sample mean for control group is  

      The  sample size  for control group is

      The standard deviation for control group is

Given that the confidence level is  95% then the level of significance can me mathematically evaluated as

Generally the degree of freedom is mathematically represented as

substituting values

Next we obtain the critical value of at a degree of freedom of  22 form the students t-distribution , the value is  

Generally the margin of error is mathematically represented as

substituting values

the 95% confidence interval for the differences of the means is mathematically represented as

substituting values

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)