Respuesta :
Answer:
[tex] (26.99-35.76) -1.98 \sqrt{\frac{4.89^2}{68} +\frac{6.43^2}{74}} =-10.659[/tex]
[tex] (26.99-35.76) +1.98 \sqrt{\frac{4.89^2}{68} +\frac{6.43^2}{74}} =-6.88[/tex]
Step-by-step explanation:
For this case we assume the following data:
7-day old: n_1 = 68 x_1 = 26.99 s_1 = 4.89
28-day old: n_2 = 74 x_2 = 35.76 s_2 = 6.43
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".
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]
The degrees of freedom are given by:
[tex] df=n_1 +n_2 -2= 68+74-2 = 140[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,140)".And we see that [tex]z_{\alpha/2}=1.98[/tex]
And replaicing we got:
[tex] (26.99-35.76) -1.98 \sqrt{\frac{4.89^2}{68} +\frac{6.43^2}{74}} =-10.659[/tex]
[tex] (26.99-35.76) +1.98 \sqrt{\frac{4.89^2}{68} +\frac{6.43^2}{74}} =-6.88[/tex]
