Answer:
The 95% confidence interval is [tex]0.88 < \mu_1 - \mu_2 < 5.12[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean for fat-blocking [tex]\= x_1 = 13.7[/tex]
The sample size for fat-blocking [tex]n = 12[/tex]
The standard deviation for fat-blocking is [tex]\sigma_1 = 2.6[/tex]
The sample mean for control group is [tex]\= x _2 = 10.7[/tex]
The sample size for control group is [tex]n_2 = 12[/tex]
The standard deviation for control group is [tex]\sigma _2 = 2.4[/tex]
Given that the confidence level is 95% then the level of significance can me mathematically evaluated as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5 \%[/tex]
[tex]\alpha =0.05[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 - 2[/tex]
substituting values
[tex]df = 12 +12 - 2[/tex]
[tex]df = 22[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of 22 form the students t-distribution , the value is
[tex]t_{\frac{\alpha }{2}, df } = 2.074[/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2}, df } * \sqrt{ \frac{\sigma^2_1 }{n_1 } + \frac{\sigma^2_2 }{n_2 } }[/tex]
substituting values
[tex]E = 2.07 4 * \sqrt{ \frac{ 2.6^2 }{12 } + \frac{2.4^2 }{12 } }[/tex]
[tex]E = 2.12[/tex]
the 95% confidence interval for the differences of the means is mathematically represented as
[tex]\= x_1 - \= x_2 - E < \mu_1 - \mu_2 < \= x_1 - \= x_2 + E[/tex]
substituting values
[tex]13.7 - 10.7 - 2.12 < \mu_1 - \mu_2 < 13.7 - 10.7 + 2.12[/tex]
[tex]0.88 < \mu_1 - \mu_2 < 5.12[/tex]
