kdenly remember that you were a smart cookie and got vaccinated against several months ago! That means that the probability that you have when you're feeling sick (before getting tested) is a mere 20%. Given d your negative test result, what is the probability that you do, in fact, DVID? ommate, Joe, has fallen down the YouTube rabbit hole. When you tell out the situation with the test he says" "Vaccine, schmaccine! The only at matters for knowing whether you have COVID is what the test says." ognitive bias is Joe exhibiting? Use the odds form of Bayes' rule to express her roommate, Ann, is also a conspiracy theorist. When you tell her COVID in
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- In a clinical study, a random sample of 540 participants agree to have their blood drawn, which is to be examined for the presence of antibodies against a certain contagious disease. It is found in 22% of the blood samples, which experimenters hope to extrapolate to the general population. From this random sample, 10 participants' blood samples are selected at random. If X is the number of samples out of the 10 who have these antibodies, what can we say about X? A. The sample size is not large enough for us to approximate X using a normal distribution B.The expected value of X is 22 C. X can be approximated using a normal distribution in lieu of a binomial distribution D. X has a sampling distribution that is normalAccording to Johns Hopkins Medicine, the healthcare providers consider a fever to be 100.4◦F. Data was acquired in a health center that observed 24 adults with COVID- 19. We want to find out if the sample mean of 104◦F body temperature of adults with COVID-19 is significantly higher than the body temperature of a person with fever. Consider the sample SD of 10.2◦F.In a Japanese study, “researchers looked at 35 peoplewith lower back pain who were enrolled in an aquatic exercise program, which included swimming and walk-ing in a pool. Almost all of the patients showed improve-ments after six months, but the researchers found that those who participated at least twice weekly showedmore significant improvement that those who wentonly once per week.”Which of the following statements is true?a) This experiment proves that swimming causes areduction in lower back pain.b) This study is a poorly designed experiment sincethere is no control group and no randomization.c) Conclusions based on this observational study aresuspect since participation in the aquatic programmay be confounded with other lifestyle behaviors thatmay cause the improvement in lower back health.d) Conclusions based on this observational study aresuspect since a sample size of 35 is too small.e) This study is not an observational study since theaquatic program is a treatment.
- Resistors labeled as 100 Ω are purchased from two different vendors. The specification for this type of resistor is that its actual resistance be within 5% of its labeled resistance. In a sample of 180 resistors from vendor A, 150 of them met the specification. In a sample of 270 resistors purchased from vendor B, 233 of them met the specification. Vendor A is the current supplier, but if the data demonstrate convincingly that a greater proportion of the resistors from vendor B meet the specification, a change will be made. a) State the appropriate null and alternate hypotheses. b) Find the P-value. c) Should a change be made?2. Consider a study where students are measured on whether they had an internship during their time at WKU (Y/N) and whether they had a job at graduation (Y/N). If we wanted to test whether having an internship was associated with having a job at graduation (i.e., internship holders were more likely to have jobs), why would the chi-square test be inappropriate for this hypothesis? How should we analyze our data?Resistors labeled as 100 Ω are purchased from two different vendors. The specification for this type of resistor is that its actual resistance be within 5% of its labeled resistance. In a sample of 180 resistors from vendor A, 149 of them met the specification. In a sample of 270 resistors purchased from vendor B, 233 of them met the specification. Vendor A is the current supplier, but if the data demonstrate convincingly that a greater proportion of the resistors from vendor B meet the specification, a change will be made. P-value?
- Ackerman and Goldsmith (2011) report that students who study from a screen (phone, tablet, or computer) tended to have lower quiz scores than students who studied the same material from printed pages. To test this finding, a professor identifies a sample of n = 16 students who used the electronic version of the course textbook and determines that this sample had an average score of M = 72.5 on the final exam. During the previous three years, the final exam scores for the general population of students taking the course averaged μ =77with a standard deviation of σ = 8 and formed a roughly normal distribution. The professor would like to use the sample to determine whether students studying from an electronic screen had exam scores that are significantly different from those for the general population. 1. Using the standard four-step procedure, conduct a two-tailed hypothesis test with αto evaluate the effect of studying from an electronic screen.An agricultural scientist tests six types of fertilizer, labeled A, B, C, D, E, and F, to determine whether any of them produces an increase in the yield of lima beans over that obtained with the current fertilizer. For fertilizer C, the increase in yield is statistically significant at the 0.05 level. For the other five, the increase is not statistically significant. The scientist concludes that the yield obtained with fertilizer C is greater than that of the current fertilizer. Explain why this conclusion is not justified.For conducting a two-tailed hypothesis test with a certain data set, using the smaller of n1-1 and n2-1 for the degrees of freedom results in df=11, and the corresponding critical values are t=+-2.201. Using the formula for the exact degrees of freedom results in df=19.063, and the corresponding critical values are t=+-2.093. How is using the critical values of t=+-2.201 more "conservative" than using the critical values of +- 2.093?
- 1. The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype. Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater evidence for indicating the correct mode of inheritance. In this problem, the true mode of inheritance is autosomal dominant, homozygous lethal. Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1. A few things of which to be mindful. 1. In the parental generation, the WT parent always has a homozygous genotype. 2. For the autosomal dominant mode of inheritance, the disease gene always homozygous for the disease allele in the parental generation. 3. For the autosomal dominant, homozygous lethal any person with the…Correlated t tests generally have more power than independent-groups t tests because the former usually have smaller standard errors of the difference between means, making the t value bigger. True FalseSuppose that a COVID-19 vaccine is 100% effective and Ro of COVID-19 is 2.5. What is the percentage of the population that needs to be fully vaccinated to reach herd immunity?