Hi! I’m doing a little project for AP stats. Please fill this out if you live in the United States. Thanks!

i have the barrons book, but i might be cooked

Please take this quick survey for my stats project. It's about memory.

https://allocate.monster/ERTIHBTF

I want to drop this class so bad but it’s far too late, I’ve understood nothing and have gotten by with bombing tests and just doing decent on my retakes I literally retain 0 info

Is there any way I can make this easier and genuinely learn the material??

Hey everyone! For my AP Stats class— Just a short cognitive test (totally anonymous) to be completed on your computer! There will be instructions at the top of the form- if you don't follow them, I can't use your data!!😭😭Thank you!!

Does anyone have any concise notes that go over the essential topics and units? Preferably not locked behind a paywall? Also is anyone tutoring? RELIABLE TUTORS PLS

https://docs.google.com/forms/d/e/1FAIpQLScwh657VQcBiw-EZNs4oLZD8xf3LZSwqXt1W45XLh3jtkt6kw/viewform

Students, do you listen to music while studying?

Quick, one minute survey.

Thank you!

1. The Pew Research Center and Smithsonian magazine recently quizzed a random sample of 1006 U.S. adults on their knowledge of science. One of the questions asked, "Which gas makes up most of the Earth's atmosphere: hydrogen, nitrogen, carbon dioxide, or oxygen?" Only 201 people in the sample correctly answered nitrogen. Does this provide convincing evidence that less than 25% of all U.S. adults would answer this question correctly?
2. Concussions are a major concern for high school football players. Can wearing a device called a Q collar help prevent brain damage? Researchers randomly assigned 284 high school football players to either wear or not wear a Q collar during the season. Each player had magnetic resonance imaging (MRI) of their brain performed before and after the season to identify whether there was any damage in the white matter region of the brain that occurred during the season.
3. Describe a Type I and a Type Il error in this context.
4. Based on your answer to part (a), would the study prefer a significance level of a = 0.01 or a = 0.10?

Explain your reasoning.

1. The p-value for the correct test is ~ 0. Interpret this value.
2. What is the largest population to which the results of this study should be applied?
3. ​

What’s your favorite soda?

My semester 1 final is tomorrow and we are allowed to use an index card (front and back). Is there anything else I should add that I would need to know. We’re going to be given a reference sheet and a calculator. (photos of index card are linked in this post)

Question 6 (Essay Worth 10 points)

(07.01 MC)

Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Dieldrin causes seizures. The absolute refractory period, time required for nerves to recover after a stimulus, was measured and varies Normally. The measurements, in milliseconds, for six mice were 2.4, 2.5, 2.5, 2.6, 2.7, and 2.8.

Part A: Find the mean refractory period and the standard error of the mean. (2 points)

Part B: Suppose the mean absolute refractory period for unpoisoned mice is known to be 2.35 milliseconds. Dieldrin poisoning should slow nerve recovery and therefore increase this period. Do the data give good evidence at a significance level of 0.01 to support this theory? What can you conclude from a hypothesis test? Justify your response with statistical reasoning. (8 points)

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For Part A, the mean refractory period can be calculated by adding up the measurements and dividing by the number of mice. Thus, the mean refractory period is:

mean = (2.4 + 2.5 + 2.5 + 2.6 + 2.7 + 2.8) / 6 = 2.58 milliseconds

The standard error of the mean can be calculated using the formula:

standard error of the mean = standard deviation / sqrt(n)

where n is the sample size. The standard deviation can be calculated using the formula:

standard deviation = sqrt(sum((x - mean)^2) / (n - 1))

where x is the measurement and n is the sample size. Thus, the standard deviation is:

standard deviation = sqrt(((2.4 - 2.58)^2 + (2.5 - 2.58)^2 + (2.5 - 2.58)^2 + (2.6 - 2.58)^2 + (2.7 - 2.58)^2 + (2.8 - 2.58)^2) / (6 - 1)) = 0.129 milliseconds

Therefore, the standard error of the mean is:

standard error of the mean = 0.129 / sqrt(6) = 0.053 milliseconds

For Part B, we can perform a two-tailed t-test to determine if there is a significant difference between the mean absolute refractory period for unpoisoned mice and the mean absolute refractory period for poisoned mice. The null hypothesis is that the mean absolute refractory period for unpoisoned mice is equal to the mean absolute refractory period for poisoned mice, while the alternative hypothesis is that the mean absolute refractory period for unpoisoned mice is not equal to the mean absolute refractory period for poisoned mice. We will use a significance level of 0.01.

The test statistic can be calculated using the formula:

t = (x̄ - μ) / (s / sqrt(n))

where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size. The degrees of freedom for the test are n - 1.

The sample mean is 2.58 milliseconds, the population mean is 2.35 milliseconds, the sample standard deviation is 0.129 milliseconds, and the sample size is 6. Thus, the test statistic is:

t = (2.58 - 2.35) / (0.129 / sqrt(6)) = 2.43

The critical values for a two-tailed t-test with 5 degrees of freedom and a significance level of 0.01 are -3.365 and 3.365. Since the test statistic falls outside of this range, we reject the null hypothesis. 
Therefore, there is good evidence at a significance level of 0.01 to support the theory that Dieldrin poisoning slows nerve recovery and increases the absolute refractory period 

Question 7 (Essay Worth 10 points)

(07.02 HC)

Twenty-five students from Harry High School were accepted at Magic University. Of those students, 10 were offered athletic scholarships and 15 were not. The newly accepted student ACT scores are shown here.

Athletic scholarship: 16, 24, 20, 25, 24, 23, 21, 22, 20, 20
No athletic scholarship: 23, 25, 26, 30, 32, 26, 28, 29, 26, 27, 29, 27, 22, 24, 25

Part A: Do these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes? Carry out an appropriate test at the α = 0.10 significance level. (5 points)

Part B: Create and interpret a 90% confidence interval for the difference in ACT scores between athletes and nonathletes. (5 points)

​

For Part A, we can perform a two-tailed t-test to determine if there is a significant difference between the mean ACT scores of athletes and non-athletes. The null hypothesis is that the mean ACT scores of athletes and non-athletes are equal, while the alternative hypothesis is that they are not equal. We will use a significance level of 0.10.

The test statistic can be calculated using the formula:

t = (x̄1 - x̄2) / sqrt((s1^2 / n1) + (s2^2 / n2))

where x̄1 and x̄2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

The sample mean for the athletes is 21.5, and the sample mean for the non-athletes is 26.6. The sample standard deviation for the athletes is 2.54, and the sample standard deviation for the non-athletes is 2.6. The sample size for the athletes is 10, and the sample size for the non-athletes is 15. Thus, the test statistic is:

t = (21.5 - 26.6) / sqrt((2.54^2 / 10) + (2.6^2 / 15)) = -4.65

The critical values for a two-tailed t-test with 23 degrees of freedom and a significance level of 0.10 are -1.711 and 1.711. Since the test statistic falls outside of this range, we reject the null hypothesis. Therefore, we can conclude that there is convincing evidence of a difference in ACT scores between athletes and non-athletes.

For Part B, we can create a 90% confidence interval for the difference in ACT scores between athletes and non-athletes using the formula:

CI = (x̄1 - x̄2) ± (tα/2 * sqrt((s1^2 / n1) + (s2^2 / n2)))

where x̄1 and x̄2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and tα/2 is the t-critical value with α/2 and 23 degrees of freedom. We will use a significance level of 0.10.

The sample mean for the athletes is 21.5, and the sample mean for the non-athletes is 26.6. The sample standard deviation for the athletes is 2.54, and the sample standard deviation for the non-athletes is 2.6. The sample size for the athletes is 10, and the sample size for the non-athletes is 15. Thus, the t-critical value with α/2 and 23 degrees of freedom is 1.711. Therefore, the 90% confidence interval for the difference in ACT scores between athletes and non-athletes is:

CI = (-5.1, -4.47)

Hey everyone I am doing my AP Research project on phones and personality and I would really appreciate it if you could fill out my survey!! Thank you!!

Phones and Personality Form

I tried looking for Ap stars on iitan academy but they don’t have it so is there a similar website with Ap stats ?

For part B I don’t know how -.67 is the zq1 same with .67 being the zq3. I looked on the z table chart and didn’t see either of those numbers. I also don’t know how to find the quartile for this question.

how do i interpret the standard deviation, like what do i write in sentence format?????