Respuesta :
Answer:
The 95% confidence interval would be given by [tex]0.281 \leq \mu_1 -\mu_2 \leq 0.529[/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_1 =9.2[/tex] represent the sample mean 1
[tex]\bar X_2 =8.8[/tex] represent the sample mean 2
n1=27 represent the sample 1 size
n2=30 represent the sample 2 size
[tex]\sigma_1 =0.3[/tex] population standard deviation for sample 1
[tex]\sigma_2 =0.1[/tex] population standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Solution to the problem
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =9.2-8.8=0.4[/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: "=-NORM.INV(0.025,0,1)".And we see that [tex]t_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281[/tex]
[tex]0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519[/tex]
So on this case the 95% confidence interval would be given by [tex]0.281 \leq \mu_1 -\mu_2 \leq 0.529[/tex]
