A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P < 0.01) than random guessing. Is it proper to conclude that these four people have ESP? No, because we would expect more than just 4 out of 500 to have ESP. No, because ESP is not a real thing. No, this could have happened just by chance given the sample size and the significance level. Yes, because they did significantly better than expected.

A political scientist claims that 38% of first-year college students characterize themselves as being “moderate” or “middle of the road” as far as their political affiliation is concerned.   Believing this claimed value is too high, you survey a random sample of 400 first-year college students and find that 120 characterize themselves as being “moderate” or “middle of the road.”    Based on this information, what will the test statistic be? Choose the answer below that is closest to what you calculate, and try not to do a lot of rounding until you get to the very end of your calculations.   1. -0.3 2. -1.2 3. -2.6 4. -3.3 5. None of the other answer options are correct because the test statistic should be positive, not negative.

According to a book published in 2011, 45% of the undergraduate students in the United States show almost no gain in learning in their first two years of college (Richard Arum et al., Academically Adrift, University of Chicago Press, Chicago, 2011). A recent sample of 1540 undergraduate students showed that this percentage is 37%. Can you reject the null hypothesis at a 2.5% significance level in favor of the alternative that the percentage of undergraduate students in the United States who show almost no gain in learning in their first two years of college is currently lower than 45%. Use both the p-value and the critical-value approaches. Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places. Zobserved = p-value = Critical value = i Hence we can conclude that the percentage of undergraduate students in the U.S. who show almost no gain in learning in their fırst two years of college is currently 45%.

Your statistics instructor claims that 60 percent of the students who take his Statistics class drive nicer cars than their classmates who didn’t take the class. For some reason that he can't quite figure out, most people don't believe him. You decide to check this out on your own. You randomly survey 84 of his past Statistics students and find that 28 feel more enrichedas a result of theclass.Test the claim, using a 0.05 significance level.zsample = -4.99 a.Identify the claim b.Identify the opposite. c.State the null hypothesis (H0). d.State the alternative hypothesis (H1) e.Find the critical value and make a sketch. f.Identify ?̂, ?,?,????.g.Compare zsample. to the critical value and state your verdict. h.Use the flow chart to state a conclusion.

This question: 5 point(s) possible In a study of 823 randomly selected medical malpractice lawsuits, it was found that 477 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped Which of the following is the hypothesis test to be conducted? OA. Ho: p=0.5 H₁: p>0.5 OC. Ho: p=0.5 H₁: p 0.5 H₁: p=0.5 OF. Ho: p0.5 H₁: p=0.5 WA (C) dismissed.

A researcher looking for evidence of extrasensory perception (ESP) tests 1000 subjects. Forty-three of these subjects do significantly better (P < 0.05) than random guessing. Forty-three seems like a lot of people, but you cannot conclude that these 43 people have ESP. Why not? The statistical significance level is not be stringent enough to conclude an ESP status. Since the tests were performed at the 5% significance level, as many as 50 subjects may have done significantly better than random guessing just by chance. The sample size was not large enough for any statistical inference. The P-value needs to be below 0.025 to be considered statistically significant given the two-tailed nature of this type of test.

I believe that the population mean number of times that adults go for a walk each week is less than 2.0. I spoke with 8 adults and found that they went on walks 3,5,2,1,0,4,0 and 5 times last week. Test this hypothesis at the 0.5 significance level.

The coach of a basketball team thinks the referee is using a weighted coin to decide which team gets the ball. He looks at the last 140 coin tosses and observes that there were 85 heads and 55 tails.  Answer the following questions. (1) What is the null hypothesis?  The observed frequency is not different than the expected frequency, meaning that the coin is biased The observed frequency is not different than the expected frequency, meaning that the coin is fair. The observed frequency is different than the expected frequency, meaning that the coin is fair. The observed frequency is different than the expected frequency, meaning that the coin is biased. (2) What is the calculated value of chi-square?  (3) How many degrees of freedom are there? (4) Assuming alpha is equal to .05, what is the critical value? (5) What is your decision?  Fail to reject the null and decide that the coin is fair. Fail to reject the null and decide that the coin is biased. Reject the null and decide that…

The dataset "nurses" is a random sample of 1,000 cases from “Health Care Provider Credential Data” from Washington State Open Data Portal. In the sample, there are 418 CNA's and 278 of the CNA's have expired credentials. Use the statistics from the nurses dataset to answer the following question: "With 90% confidence, what percent of all CNA's on Washington State's healthcare provider list are expired?"