Respuesta :
Answer:
The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).
Step-by-step explanation:
Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes 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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
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.
A simple random sample of 22 students who attended a review session was selected, and the mean grade on test 7 for this sample of 22 students was 85 with a standard deviation of 9.8.
This means that:
[tex]\mu_A = 85, s_A = \frac{9.8}{\sqrt{22}} = 2.09[/tex]
An independent simple random sample of 47 students who did not attend a review session was selected, and the mean grade on test 7 was 77 with a standard deviation of 10.6.
This means that:
[tex]\mu_N = 77, s_N = \frac{10.6}{\sqrt{47}} = 1.55[/tex]
Distribution of the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session.
Mean:
[tex]\mu = \mu_A - \mu_N = 85 - 77 = 8[/tex]
Standard deviation:
[tex]s = \sqrt{s_A^2 + s_B^2} = \sqrt{2.09^2 + 1.55^2} = 2.6[/tex]
Confidence interval:
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 Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
Now, find the margin of error M as such
[tex]M = zs[/tex]
So
[tex]M = 1.645*2.6 = 4.3[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 8 - 4.3 = 3.7
The upper end of the interval is the sample mean added to M. So it is 8 + 4.3 = 12.3
The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).
