Respuesta :
Answer:
[tex] (69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908[/tex]
[tex] (69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308[/tex]
So then the confidence interval for the difference of means is given by:
[tex]-8.908 \leq \mu_M -\mu_F \leq 5.308[/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]\bar X_{M}= 69.8[/tex] represent the mean for the age of grandmother
[tex]\bar X_{F}= 71.6[/tex] represent the mean for the age of grandfather
[tex]s_{M}= 8.38[/tex] represent the sample deviation for the age of grandmother
[tex]s_{F}= 6.65[/tex] represent the sample deviation for the age of grandfather
[tex]n_M = n_F= 10[/tex]
Solution to the problem
For this case the confidence interval is given by:
[tex] (\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}} [/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_M +n_F-2=10+10-2=18[/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(0.025,18)".And we see that [tex]t_{\alpha/2}=2.101[/tex]
And replacing we got:
[tex] (69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908[/tex]
[tex] (69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308[/tex]
So then the confidence interval for the difference of means is given by:
[tex]-8.908 \leq \mu_M -\mu_F \leq 5.308[/tex]
