The first pic is the question. Since some tutors have been solving the problems wrong recently (no judgment), I’ve added a sample question to solve problem. Sample QuestionHockey's Highest Scorers The number of points held by a sample of the NHL's highest scorers for both the Eastern Conference and the Western Conference is shown below. Ata = 0.01, can it be concluded that there is a difference in means based on these data? Assume the variables are normally distributed and the variances are unequal.Eastern Conference Western Conference77 37 72 5775 78 708862 59 61 9866556164596270 7166687682Send data to ExcelUse u, for the mean score for the Eastern Conference.(a) State the hypotheses and identify the claim with the correct hypothesis.(b) Find the critical value(s).(c) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim with the correct hypothesis.The null hypothesis IH, is the statement that there is no difference between the means. This is equivalent to u, -H,.The alternative hypothesis H is the statement that there is a difference between the means. This is equivalent to u, #µ,.The problem asks if "there is a difference in means based on these data." Hence, the claim is the alternative hypothesis H,.(b) Find the critical value(s).For this problem, n, = 13 and n, = 11. The degrees of freedom are the smaller of 13 –1= 12 or 11-1= 10. Hence, d.f. = 10.From OThe t Distribution Table, for a two-tailed test with a = 0.01 and d.f. = 10, the critical values are 13.169.(c) Compute the test value.Use a calculator/computer/formulas in Chapter 3 to compute the sample statistics for each conference. The results are shown below (Round the sample means and standard deviationsto two decimal places):Eastern Conference Western ConferenceSample mean71.9263.55Sample standard deviation12.0611.25Sample size1311Using the formula for the i test-for testing the difference between two means-independent samples, compute the test value:(X, - X2)- (H, – 4-)2.-+-(71.92 – 63.55)-012.06 11.25?1311= 1.757Hence, the test value rounded to three decimal places is t= 1.757.(d) Make the decision.Since the test value does not fall in the critical region, do not reject the null hypothesis. Hockey's Highest Scorers The number of points held by a sample of the NHL's highest scorers for both theEastern Conference and the Western Conference is shown below. At a=0.10, can it be concluded that there is adifference in means based on these data? Assume the variables are normally distributed and the variances areunequal.Eastern ConferenceWestern Conference61896288776472668359609859583757757870626655616461596266Send data to ExcelUse u, for the mean score for the Eastern Conference.Part: 0 / 5Part 1 of 5State the hypotheses and identify the claim with the correct hypothesis.Ho :|(Choose one) ▼H :(Choose one)This hypothesis test is a(Choose one) ▼test.

Answered: Image /qna-images/answer/ea153cb0-1fc8-4ccd-9e0f-1231a70da00d.jpg

Hockey's Highest Scorers The number of points held by a sample of the NHL's highest scorers for both the Eastern Conference and the Western Conference is shown below. At a=0.10, can it be concluded that there is a difference in means based on these data? Assume the variables are normally distributed and the variances are unequal. Eastern Conference Western Conference 61 89 62 88 77 64 72 66 83 59 60 98 59 58 37 57 75 78 70 62 66 55 61 64 61 59 62 66 Send data to Excel Use u, for the mean score for the Eastern Conference. Part: 0 / 5 Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. Ho : |(Choose one) ▼ H : (Choose one) ▼ This hypothesis test is a(Choose one) ▼ test.

Hockey's Highest Scorers The number of points held by a sample of the NHL's highest scorers for both the Eastern Conference and the Western Conference is shown below. At a=0.10, can it be concluded that there is a difference in means based on these data? Assume the variables are normally distributed and the variances are unequal. Eastern Conference Western Conference 61 89 62 88 77 64 72 66 83 59 60 98 59 58 37 57 75 78 70 62 66 55 61 64 61 59 62 66 Send data to Excel Use u, for the mean score for the Eastern Conference. Part: 0 / 5 Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. Ho : |(Choose one) ▼ H : (Choose one) ▼ This hypothesis test is a(Choose one) ▼ test.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

14) The average retail price for bananas in 2017 was 56.0 cents per pound. Recently, a random sample of 15 food stores resulted in a mean of 57.4 cents per pound with a standard deviation of 2.5 cents. Using the Minitab printout below, test the claim that the price of bananas has increased at the 0.01 significance level? (Show all six steps.)
A school superintendent wanted to know whether there were differences between the high schools in her district with regards to the proportion of students who do not live with their biological parents. She surveyed 200 students from each of the two high schools – East and West. The proportion of students from the survey at East who do not live with their biological parents is 3.4%. The proportion of students from the survey at East who do not live with their biological parents is 4.2%. Using a 95% confidence interval (2SD or Theory based), can the superintendent conclude that there is a difference between East and West with respect to the proportion of students who do not live with their biological parents. Group of answer choices Yes she can conclude there is a difference because the confidence intervals overlap. No she cannot conclude there is a difference because the confidence intervals overlap. Yes she can conclude there is a difference because the confidence intervals don’t…
what are some tips to study for the FSA test like the reading and math as well.
Help please with P-value. How to compute in excel or statcrunch would be helpful.
In a multiple-choice test, there are three possible answers to each question. There are 10 questions in the exam. One point is given for the correct answer but nothing is deducted for the wrong answer. To pass the exam you must get at least 5 points. Sigmundur is very bad at the study material and therefore decides what the correct answer is in each question. What are the chances that Sigmundur will pass the exam? 1. about 2%. 2. about 7%. 3. about 14%. 4. about 21%.
Solve number 5 and explain
When taking a 12 question multiple choice test, where each question has 3 possible answers, it would be or more questions correct by guessing alone? unusual to get Give your answer in the box above as a whole number.
Thomas scored at the 45th percentile on the midterm. Which interpretation of his score is the best? Group of answer choices 45% of the people tested exceeded his score. Thomas had a better score than 45% of the people tested. Thomas got 45% of the items correct. More than one of the above is appropriate.
A survey found that 31% of Americans watch the fireworks on the 4th of July from their TV. If 20 Americans are surveyed, would it be unusual to have less than 8 watch the fireworks from their TV? Explain.
Explain about Other Multiple Comparison Tests.
At Cityville High School, 36% of the students participate in sports, 36% of the students participate in academic clubs, and 13% of the students participate in both sports and academic clubs. Find the proportion of students who participate in either sports or academic clubs.
Please teach me on how to solve this problem. C is wrong, maybe B, because it's talking about "sample ", and definitely "36students " are just samples, not population. Thanks so much.