Respuesta :
Answer:
a) [tex]\bar X_1 -\bar X_2 =75-68=7[/tex]
b) [tex]SE= \sqrt{\frac{6^2}{36}+\frac{7^}{49}}=1.414[/tex]
On this case the 95% confidence interval would be given by [tex]4.186 \leq \mu_1 -\mu_2 \leq 9.814[/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 =75[/tex] represent the sample mean 1
[tex]\bar X_2 =68[/tex] represent the sample mean 2
n1=36 represent the sample 1 size
n2=49 represent the sample 2 size
[tex]s_1 =6[/tex] sample standard deviation for sample 1
[tex]s_2 =7[/tex] sample standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Part a
The confidence interval for the difference of means is given by the following formula:
[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] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =75-68=7[/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_1 +n_2 -1=36+49-2=83[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,83)".And we see that [tex]t_{\alpha/2}=1.99[/tex]
Part b
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]
And replacing we have:
[tex]SE=\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=1.414[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]7-1.99\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=4.186[/tex]
[tex]7+1.99\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=9.814[/tex]
So on this case the 95% confidence interval would be given by [tex]4.186 \leq \mu_1 -\mu_2 \leq 9.814[/tex]
Using the Central Limit Theorem, it is found that:
a) The estimate is of 7 units.
b) The standard error is of 1.414 units.
Item a:
- This year, the mean is [tex]\mu_1 = 75[/tex]
- Last year, the mean was [tex]\mu_2 = 68[/tex]
By the Central Limit Theorem, the estimate for the mean of the difference distribution is the difference of the means, thus:
[tex]\mu = \mu_1 - \mu_2 = 75 - 68 = 7[/tex]
The estimate is of 7 units.
Item b:
By the Central Limit Theorem, the standard error for each sample is the standard deviation for the sample divided by the sample size, thus:
[tex]s_1 = \frac{6}{\sqrt{36}} = 1[/tex]
[tex]s_2 = \frac{7}{\sqrt{49}} = 1[/tex]
The standard error of the difference is the square root of the sum of the variances, thus:
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2} = 1.414[/tex].
The standard error is of 1.414 units.
A similar problem is given at https://brainly.com/question/23274490
