Mathematics, 05.03.2021 21:50 madvitch6111

Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 453 randomly selected students in Florida; in this sample, the mean is 1013 and the standard deviation (5) is 108. a. Construct a 95 % confidence interval for the average test score for Florida students.
b. Is there statistically significant evidence that Florida students perform differently than other students in the United States?
c. Another 503 students are selected at random from Florida. They are given a 3-hour preparation course before the test is administered. Their average test score is 1019 with a standard deviation of 95.
i. Construct a 95% confidence interval for the change in average test score associated with the prep course.
ii. Is there statistically significant evidence that the prep course helped?

d. The original 453 students are given the prep course and then are asked to take the test a second time. The average change in their test scores is 9 points, and the standard deviation of the change is 60 points.
i. Construct a 95% confidence interval for the change in average test scores.
ii. Is there statistically significant evidence that students will perform better on their second attempt after taking the prep course?
iii. Students may have performed better in their second attempt because of the prep course or because they gained test- taking experience in their first attempt. Describe an experiment that would quantify these two effects.

Answers: 3

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Answer from: lizisapenguin

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, $\bar{x}$ = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

$CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}$

At 95% confidence level, z = 1.96, therefore, we have;

$CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}$

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, $\bar{x}$ = 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;

$C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore, we have;

$C.I. = \left (1013- 1019 \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}$

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, $\bar{x}_{1}- \bar{x}_{2}$ = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = $\bar{x}_{1}- \bar{x}_{2}$ + 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

Answer from: Quest

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.