Respuesta :
Answer:
a. 2
b. The 90% confidence interval for the difference between the two population means is (1.02, 2.98).
c. The 95% confidence interval for the difference between the two population means is (0.83, 3.17).
Step-by-step explanation:
Before solving this question, we need to understand the central limit theorem and the subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Sample 1:
[tex]\mu_1 = 13.6, s_1 = \frac{2.2}{\sqrt{50}} = 0.3111[/tex]
Sample 2:
[tex]\mu_2 = 11.6, s_2 = \frac{3}{\sqrt{35}} = 0.5071[/tex]
Distribution of the difference:
[tex]\mu = \mu_1 - \mu_2 = 13.6 - 11.6 = 2[/tex]
[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.3111^2+0.5071^2} = 0.595[/tex]
a. What is the point estimate of the difference between the two population means?
Sample difference, so [tex]\mu = 2[/tex]
b. Provide a 90% confidence interval for the difference between the two population means.
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
The margin of error is:
[tex]M = zs = 1.645(0.595) = 0.98[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 2 - 0.98 = 1.02
The upper end of the interval is the sample mean added to M. So it is 2 + 0.98 = 2.98
The 90% confidence interval for the difference between the two population means is (1.02, 2.98).
c. Provide a 95% confidence interval for the difference between the two population means.
Following the same logic as b., we have that [tex]Z = 1.96[/tex]. So
[tex]M = zs = 1.96(0.595) = 1.17[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 2 - 1.17 = 0.83
The upper end of the interval is the sample mean added to M. So it is 2 + 1.17 = 3.17
The 95% confidence interval for the difference between the two population means is (0.83, 3.17).
