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Sagot :
Answer:
a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544
b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States
c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390
ii. There are no statistical significant evidence that the prep course helped
d. i. The 95% confidence interval for the change in average test scores is 3.47467 < μ₁ - μ₂ < 14.52533
ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course
iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience
Step-by-step explanation:
The mean of the standardized test = 1,000
The number of students test to which the test is administered = 453 students
The mean score of the sample of students, [tex]\bar{x}[/tex] = 1013
The standard deviation of the sample, s = 108
a. The 95% confidence interval is given as follows;
[tex]CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}[/tex]
At 95% confidence level, z = 1.96, therefore, we have;
[tex]CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}[/tex]
Therefore, we have;
1,022.94559 < μ < 1,003.0544
b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States
c. The parameters of the students taking the test are;
The number of students, n = 503
The number of hours preparation the students are given, t = 3 hours
The average test score of the student, [tex]\bar{x}[/tex] = 1019
The number of test scores of the student, s = 95
At 95% confidence level, z = 1.96, therefore, we have;
The confidence interval, C.I., for the difference in mean is given as follows;
[tex]C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}[/tex]
Therefore, we have;
[tex]C.I. = \left (1013- 1019 \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}[/tex]
Which gives;
-18.955390 < μ₁ - μ₂ < 6.955390
ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped
d. The given parameters are;
The number of students taking the test = The original 453 students
The average change in the test scores, [tex]\bar{x}_{1}- \bar{x}_{2}[/tex] = 9 points
The standard deviation of the change, Δs = 60 points
Therefore, we have;
C.I. = [tex]\bar{x}_{1}- \bar{x}_{2}[/tex] + 1.96 × Δs/√n
∴ C.I. = 9 ± 1.96 × 60/√(453)
i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533
ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course
iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience
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