Find tα/2 when n=12 for the 95% confidence interval for the mean.
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- The antibody production of 12 male red-winged blackbirds before and after receiving testosterone implants was compared. The units for antibody levels were natural log (10-3 optical density) per minute (In(mOD/min)). The mean change in antibody production was d = 0.056, and the standard deviation was sd = 0.225 If you were assigned the task of repeating this experiment, and wanted to ensure that you could detect a mean change of 0.02 units with a probability of 0.8, then what sample size would you use? Since we are calculating n for a study of individuals, answers should be rounded up to the next whole number.You have developed an exam to measure knowledge of biology and administered it to a large number of people. On this exam, u= 85 and o = 10. You standardize the distribution so that the new mean and standard deviation are u = 50 and o = 5 If someone takes the exam and receives a score of 73 (before the score is transformed), what will this score be once it is transformed for the new (standardized) distribution?A man measures his heart rate before using a treadmill and then after walking ona treadmill for 10 minutes on 7 separate days. His mean heart rate at baseline and10 minutes after treadmill walking is 85 and 93 beats per minute (bpm), respectively. The mean change from baseline to 10 minutes is 8 bpm with a standarddeviation of 6 bpm.(a) What test can we use to compare pre- and post-treadmill heart rate?(b) Implement the test in 2a, and report a two-tailedp-value.(c) Provide a 90% confidence interval (CI) for the mean change in heart rate afterusing the treadmill for 10 minutes.(d) What is your overall conclusion concerning the data? this is a review question of ch8 bernard rosner. the solution is not provided at bartleby for this review question
- find the are under the normal curve between Z-scores of - 2.06 and 1,91 -1.82 and 2.68 -0.57 and -1.42 Show and explain complete solutionA standardized exam’s score are normally distributed. In a recent year, the mean test score was 1477 and the standard was 315. The test scores of four students selected at random 1860 1230, 2160, and 1360. Find the Z- score that correspond to each value and determine wether any of the values are unusual. The Z-score for 1860 is-The Z-score for 1230 is-The Z-score for 2160 is-The Z-score for 1360is-A researcher collected sample data for 10middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 192.5, with a standard deviation of 9. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 95% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 95% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. Lower Limit___ Upper Limit___
- Let x be a random variable that represents the pH of arterial plasma (l.e., acidity of the blood). For healthy adults, the mean of the x distribution is = 7.4.t A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.8 with sample standard deviations = 3.3. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. A USE SALT (a) What is the level of significance? State the null and alternate hypotheses. O Hg: u + 7.4; H,: = 7.4 O Hg: H = 7.4; H,: H > 7.4 O H,: H = 7.4; H,i 7.4; H,: H = 7.4 O H: = 7.4; H,: H# 7.4 (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The Student's t, since the sample size is large a is known. O The standard normal, since the sample size is large and a is known. O The standard normal,…Assume that a high school principal believes that high number of absences hinders the students' ability to perform on standardized tests. He randomly selects a sample of 9 students who had 5 or more absences in the school year. The average ACT Composite score for this sample is X = 19 with s = 4.9. ACT Composite scores are normally distributed with u = 21.0. Do the data support the claim that the high number of absences hinders the students' ability to perform on standardized tests, using a = .05? Please show the four-step hypothesis test. %3DIt takes an average of 14.5 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 70 injured patients to immediately tell the truth about the injury and noticed that they averaged 16.3 minutes for their blood to begin clotting after their injury. Their standard deviation was 4.75 minutes. What can be concluded at the the α = 0.10 level of significance? a. For this study, we should use [Select an answer b. The null and alternative hypotheses would be: Ho: ? Select an answer H₁: ? Select an answer ✓ c. The test statistic ? = (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? ✓ a f. Based on this, we should [Select an answer the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest the populaton mean is significantly greater than 14.5 at a = 0.10, so…
- Hi. I've been trying to figure out my public health stats homework for a few hours and decided to try to get some help here. I've written down the homework questions and provided the data needed from the table. I can't seem to figure out what is being asked of me. Thanks for any help. QUESTIONS: Use the information in table 5 to answer the following questions. A. Suppose moms of both children with asthma and moms who have children without asthma misreported infant supplementation of cod liver oil. Twenty five percent of the moms that reported no cod liver oil supplementation actually sometimes gave their infants cod liver oil. 1. Create the corrected 2X3 table (show your work). 2. Recalculate all of the incidence proportions. 3. Recalculate the relative risks. 4. Did the misclassification under- or over- estimate the effect of cod liver oil on the development of asthma reported in the manuscript? 5. Is this an example of non-differential or differential misclassification? Why? B.…A cell phone manufacturer has hired you to estimate the population mean of the battery lifetimes for all phones of their latest model. You decide to measure battery lifetime by recording the time it takes for the battery charge to run out while a tester is streaming videos on the phones continuously. Then you select a random sample of 45 cell phones of the manufacturer's latest model and record their battery lifetimes. Assume that the population standard deviation of the battery lifetimes for that cell phone model (using your method of measurement) is 2.42 hours. Based on your sample, follow the steps below to construct a 95% confidence interval for the population mean of the battery lifetimes for all phones of the manufacturer's latest model. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results from your random sample of 45 phones of the manufacturer's latest model. Take Sample Sample size: 0 Number of phones Point estimate: 0 45 Sample mean…Exam scores in a large class are normally distributed with a mean of 55 and standard deviation of 10. The teacher decides to boost the scores by adding 20 points to everyone’s score. Elizabeth’s original score on the exam was 75, so her new score is 95. The z-score corresponding to her new score (relative to the other new scores) will be:
