Helpppp please i need to turn this in asap

sammy [17]

Hello, I just did this lesson and I want to say the answer is 17%. I’m not 100% sure on that answer and I wanted to give you a heads up on that. I hope this is right!

6 0

3 years ago

Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45

vovikov84 [41]

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, $\bar{x}$ = 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

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

5 0

3 years ago

Assume the sample variances to be continuous measurements. Find the probability that a random sample of 25observations, from a n

konstantin123 [22]

Answer:

a) $P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

b) $P(13.848$ Step-by-step explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

For this case we assume that the sample variance is given by $S^2$ and we select a random sample of size n from a normal population with a population variance $\sigma^2$ . And we define the following statistic:

$T = \frac{(n-1) S^2}{\sigma^2}$

And the distribution for this statistic is $T \sim \chi^2_{n-1}$

For this case we know that n =25 and $\sigma^2 = 6$ so then our statistic would be given by:

$\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2$

With 25-1 =24 degrees of freedom.

Solution to the problem

Part a

For this case we want this probability:

$P(S^2 > 9.1)$

And we can multiply the inequality by 4 on both sides and we got:

$P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"

Part b

For this case we want this probability:

$P(3.462 < S^2$

If we multiply the inequality by 4 on all the terms we got:

$P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2$ And we can find this probability like this: $P(13.848$ And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"

8 0

3 years ago

Which equation will Alex want to use to determine the total amount of

Ghella [55]

Answer:

f= 5x

Step-by-step explanation:

x is the amount of food you want over 5 years so you would have to times it to get the whole amount eaten after 5 years.

6 0

3 years ago

Last year scott ate 84 cups of ice cream.how many quarts did he eat?

harina [27]

Answer:

21 quarts

Step-by-step explanation:

1 cup = 1/4 quart

84 cups = 1/4 x 84 quarts

84/4 = 21

84 cups = 21 quarts.

4 0

3 years ago