ACT Scores A survey of 1000 students nationwide showed a mean ACT score of 21.4. A survey of 500 tllinois scores showed a mean of 21.3. If the population standard deviation in each case is 3, can we conclude the national average is greater than the Illinois average? Use a = 0.01 and use u, for the nationwide mean ACT Score. Part: 0 /5 Part 1 of 5 (a) State the hypotheses and identify the claim. Ho: (Choose one) D=0 H: (Choose one) ▼ H2 test. This hypothesis test is a (Choose one) v Submit Assignmer Save For Later Check A Righe Resved Tems of Use Pacy Skip Part 2021 McGrawHl Edu

ACT Scores A survey of 1000 students nationwide showed a mean ACT score of 21.4. A survey of 500 tllinois scores showed a mean of 21.3. If the population standard deviation in each case is 3, can we conclude the national average is greater than the Illinois average? Use a = 0.01 and use u, for the nationwide mean ACT Score. Part: 0 /5 Part 1 of 5 (a) State the hypotheses and identify the claim. Ho: (Choose one) D=0 H: (Choose one) ▼ H2 test. This hypothesis test is a (Choose one) v Submit Assignmer Save For Later Check A Righe Resved Tems of Use Pacy Skip Part 2021 McGrawHl Edu

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Description: ACT Scores A survey of 1000 students nationwide showed a mean ACT score of 21.4. A survey of 500 fllinois scores showed a mean of 21.3. If the populationstandard deviation in each case is 3, can we conclude the national average is greater than the I

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Standardized tests: In a particular year, the mean score on the ACT test was 19.7 and the standard deviation was 4.2. The mean score on the SAT mathematics test was 518 and the standard deviation was 103. The distributions of both scores were approximately bell-shaped. Round the answers to at least two decimal places. Find the z-score for an ACT score of 23. Find the z-score for a SAT score of 466.
Part d) where does the 0.675 come from?
Dice. When one die is rolled, the number of spots show-ing has a mean of 3.5 and a standard deviation of 1.7. Suppose you roll 10 dice. What’s the approximate prob-ability that your total is between 30 and 40 (that is, the average for the 10 dice is between 3 and 4)? Specify themodel you use and the assumptions and conditions thatjustify your approach.
I Professor Brown administered an achievement test, which as he remembers from years of experience, yielded a population mean score of 85. If his class of 25 students obtained a mean score of 89 with standard deviation score of 14, can the professor claim that his class is a superior class? Assume that the mean score is approximately normally distributed. Use a = 0.01
Given: u = 500 ( Population mean score for a Standardized Biology exam. 0 = 30 (Population mean score for a Standardized Biology exam. %3D Eighty-one (81) scores were selected at random. The mean, X-bar, was calculated. 3. Find P( X-bar 510). . 5. Find P( 480 < X-bar < 520) · Given : The average test score (u) in 2020 of a Standardized Accounting Exam was 550 with a population standard deviation (o) of 45. 6. . Find Lowest Test Score x which is in the 95" Percentile. 7. Find Lowest Test Score x which is in the 75" Percentile. 8. Find Lowest Test Score x which is in the 25" Percentile. 75t 25 that an Independent Education Research team desires to estimate the populatie The sample mean was 3.15.
Just answer: According to the P value, what is the mean difference statistically significant? Based on the results, can you reject the null hypothesis? Is the entire staff operating at a good level?
Part b and c
A quiz related to digital knowledge was conducted. The quiz had 10 questions and covered topics such as the purpose of browser cookies, phishing scams, and privacy policies. The survey was given to 50 people. The mean score is 4.1 with a standard deviation of 2.6. We want to know if the data provide evidence that the mean score in the population is lower than 5. We will use alpha = 0.05 to make our decision. part a. Is this a one-sided or two-sided test? part b. What is the standard error for this problem? part c. How many degrees of freedom do we have for this problem? part d. What is the test statistic, t, for this problem? Based on your test statistic and degrees of freedom, the p-value will be between what two values? (If you cannot find the exact df in the table, round down to be safe. So if you have 38 df, round down to 30df.) Group of answer choices The p-value is between .1 and .05 The p-value is between .05 and .025 The p-value is between .025 and .01 The p-value…
The mean number of eggs per person eaten in the United States is 252 and the standard deviation is 68. Do college students eat more than the average? The 45 college students surveyed averaged 281 eggs per person. What can be concluded at the 0.05 level of significance? H0: = 252 Ha: _____________ 252 Test statistic: __________ p-Value =__________ Reject/Fail to reject H0?______________ Conclusion: There is____________ evidence to make the conclusion that the population mean number of eggs consumed college students per year is more than 252 eggs per student per year. p-Value Interpretation: If the mean number of eggs consumed per year for college students is equal to_________ and if another study was done with a new randomly selected group of 45 college students, then there is a ____________ percent chance that the average number of eggs consumed per year for this new sample would be greater than_____________ . Level of significance…
EXPLAIN STEP 1, 2, AND 3please do this typewritten and completely so I will upvote. Thank you and much appreciated for the help.
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▼ Part 1 of 4 Londell, an accomplished marathon runner, claims that the standard deviation for his time to run a 26.2 mile marathon is less than 11.1 minutes. To test his claim, Londell looks up 17 of his race times (in minutes): 238.6, 240, 231.4, 241.4, 251.9, 230.4, 232.1, 235.7, 234.6, 242.9, 240.5, 237.6, 242.2, 245.7, 235.7, 240.2, 233.9 Perform a hypothesis test using a 1% level of significance to test Londell's claim. Step 1: State the null and alternative hypotheses. Ho:o 11.1 Ha:o