Question

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.

Answer 1

The 95% confidence interval is
`0.88 < \mu_1 - \mu_2 < 5.12`

Explanation:

From the question we are told that

The sample mean for fat-blocking
`\= x_1 = 13.7`

The sample size for fat-blocking
`n = 12`

The standard deviation for fat-blocking is
`\sigma_1 = 2.6`

The sample mean for control group is
`\= x _2 = 10.7`

The sample size for control group is
`n_2 = 12`

The standard deviation for control group is
`\sigma _2 = 2.4`

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

`\alpha = 100 - 95`

`\alpha = 5 \%`

`\alpha =0.05`

Generally the degree of freedom is mathematically represented as

`df = n_1 + n_2 - 2`

substituting values

`df = 12 +12 - 2`

`df = 22`

Next we obtain the critical value of
`(\alpha )/(2)` at a degree of freedom of 22 form the students t-distribution , the value is

`t_{(\alpha )/(2), df } = 2.074`

Generally the margin of error is mathematically represented as

`E = t_{(\alpha )/(2), df } * \sqrt{ (\sigma^2_1 )/(n_1 ) + (\sigma^2_2 )/(n_2 ) }`

substituting values

`E = 2.07 4 * \sqrt{ ( 2.6^2 )/(12 ) + (2.4^2 )/(12 ) }`

`E = 2.12`

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

`\= x_1 - \= x_2 - E < \mu_1 - \mu_2 < \= x_1 - \= x_2 + E`

substituting values

`13.7 - 10.7 - 2.12 < \mu_1 - \mu_2 < 13.7 - 10.7 + 2.12`

`0.88 < \mu_1 - \mu_2 < 5.12`