Respuesta :
Answer:
[tex] (16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445[/tex]
[tex] (16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]n_1 = 19 [/tex] sample size 1
[tex]n_2 = 19 [/tex] sample size 2
[tex]\\bar X_1 = 16.2[/tex] sample mean for group 1
[tex]\\bar X_2 = 14.2[/tex] sample mean for group 2
[tex]s_1 = 4.5[/tex] sample deviation for group 1
[tex]s_2 = 3.1[/tex] sample deviation for group 2
Solution to the problem
For this case the confidence interval is given by:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}[/tex]
And the degrees of freedom are given by:
[tex] df = n_1 +n_2 -2 = 19+19 -2= 36[/tex]
We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and [tex]\alpha/2 = 0.025[/tex] if we find a critical value in the t distribution with 36 degrees of freedom we got:
[tex] t_{cric} =2.03[/tex]
And replacing we got:
[tex] (16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445[/tex]
[tex] (16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645[/tex]
