A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor m
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We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.
Answer:
A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.
False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.
B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.
Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.
C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.
False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.
Step-by-step explanation:
Notation and previous concepts
represent the sample after the preparation
represent the sample without preparation
represent the mean sample after preparation
represent the mean sample without preparation
represent the sample deviation after preparation
represent the sample deviation without preparation
represent the significance level
Confidence =90% or 0.90
The confidence interval for the difference of means is given by the following formula:
(1)
The point of estimate for
The appropiate degrees of freedom are
Since the Confidence is 0.90 or 90%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)
The standard error is given by the following formula:
After replace in the formula for the confidence interval we got this:
And we need to interpret this result:
A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.
False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.
B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.
Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.
C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.
False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.