3. The time until a certain pain medication reach es maximum level in the bloodstream has been measured in 22 patients. The sample aver age time was 27 minutes and the sample standard deviation was 4 minutes. Assume that the time to reach the maximum level is norm ally dis- tributed. (a) Compute a 90% confidence interval for the mean time until the maximum level of medication is reached. Compute a 95% confidence interval for the mean time until the maximum level (b) of medication is reached. Compute a 99% confidence interval for the mean time until the maximum level (c) of medication is reached.
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- What is meant by the sample space of an experiment?Suppose a retailer claims that the average wait time for a customer on its support line is 185 seconds. A random sample of 53 customers had an average wait time of 177 seconds. Assume the population standard deviation for wait time is 53 seconds. Using a 95% confidence interval, does this sample support the retailers claim? A) Yes, because the retailers claim is beteeen the lower limit of __ seconds and the upper limit of __ seconds for the mean wait time. B) No, because the retailers claim is not between the lower limit of __ seconds and the upper limit of __ seconds for the mean wait time. Please help!1) A Gallup poll of 1015 randomly selected adults found that 69% would oppose a law that limited the size of soft drinks and other sugary beverages served in restaurants. Calculate and interpret a 95% confidence interval for the proportion of all adults that would oppose a law limiting the size of soft drinks.
- A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain areawho favor moving the broadcast of the late weeknight news to an hour earlier than it is currently. Initially, theconfidence interval will be created using a simple random sample of 9,000 viewers in the area. Assuming that thesample proportion does not change, what would be the relationship between the width of the original confidenceinterval and the width of a second 90 percent confidence interval that is created based on a sample of only 1,000viewers in the area?(A) The second confidence interval would be 9 times as wide as the original confidence interval.(B) The second confidence interval would be 3 times as wide as the original confidence interval.(C) The width of the second confidence interval would be equal to the width of the original confidence interval.(D) The second confidence interval would be 1/3 as wide as the original confidence interval.(E) The second…In a study of academic procrastination, the authors of a paper reported that for a sample of 431 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.54 hours and the standard deviation of study times was 3.90 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university. (a) Construct a 95% confidence interval to estimate ?, the mean time spent studying for the final exam for students taking introductory psychology at this university. (Round your answers to three decimal places.)( , )(b) The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam. n = 431 x = 43.88 s = 21.66 Construct a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior…4) The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A 90% confidence interval that results from examing 607 customers in Taco Bell's drive-through has a lower bound of 161.6 seconds and an upper bound of 164.7 seconds. What does this mean?
- The Okeechobee County Economic Development Board needs to estimate the proportion of all county residents who have earned at least a bachelor's degree. A random sample conducted by the Okeechobee County Economic Development Board found that 10.5% of all county residents held at least a bachelor's degree.Find three different confidence intervals - one with sample size 180, one with sample size 451, and one with sample size 621. Assume that 10.5% of the county residents in each sample have earned at least a bachelor's degree. Notice how the sample size affects the margin of error and the width of the interval.Report confidence interval solutions using interval notation. Report all solutions in percent form, rounded to two decimal places, if necessary. When n=180n=180, the margin of error for a 95% confidence interval is given by .When n=180n=180, a 95% confidence interval is given by . When n=451n=451, the margin of error for a 95% confidence interval is given by .When n=451n=451, a…1) An electronic manufacturer produces lampshades with a life span of approximately normally distributed with a standard deviation of 35 hours. a) If a sample of 50 bulbs has an average life span of 835 hours, construct a 95% confidence interval for the average life span of all the lampshades produced. Interpret the result. b) What must be the sample size if the manufacturer wishes to be 99% confident that the estimate is within 10 hours of the correct mean? Interpret the result.According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this percent is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Suppose we want to calculate a 99% confidence interval for the difference between the proportions of Californians and Oregonians who are sleepdeprived. 1. State the critical value for the interval (3 decimal places): 2. State the difference between the sample proportions, POregon-PCalifornia (3 decimal places): 3. State the standard error of the difference between the proportions (3 decimal places): 4. State the confidence interval being sure to use proper notation (3 decimal places each):
- A survey of entry level corporate analysts in New York City asked, approximately how many hours per week do you work? A) If the population standard deviation is known to be 4.7 hours, and a random sample of 35 analysts yielded a mean of 50.5 hours, then construct a 90% confidence interval for the mean number of hours worked by entry level analysts. B) In the above confidence interval, suppose that we want a margin of error of +/- 1 hour. What sample size would be needed to achieve this? C) As mentioned in class, it is unlikely that the population standard deviation is known. Suppose instead that the sample standard deviation was 4.7 hours. Using the other values from part (a), form a 90% confidence interval for the mean number of hours worked by entry level analysts.A research report on a prospective study states that: “A comparison of the 90 normal patients with the 50 patients with hypotension shows that only 3.33% of the former group died as compared to 30% of the latter.” Put these data into a 2x2 contingency table. E- Hypotension Healthy Total Death 15 3 18 No death 35 87 122 Total 50 90 140 Calculate the sample risk difference, construct a 95% confidence interval, and provide an interpretation. Calculate the sample relative risk, construct a 95% confidence interval, and provide an interpretation. Calculate the sample odds ratio, construct a 95% confidence interval, and provide an interpretation. Test the significance of blood pressure as a prognostic sign using Pearson’s Chi-Squared Test. Be sure to provide the following: Define the parameter(s) of interest. State the null and alternative hypotheses. State the significance level. State what type of test you will perform…A healthcare information company is interested in estimating the average charge for a standard patient visit to a chiropractor in Maryland, after applying the discount negotiated with a large HMO plan. Data is collected from 16 randomly selected chiropractic practices in Maryland, and the following are some summary statistics:- Mean charge: 25.50 USD- SD of charges: 2.10 USDAssuming the charge data is normally distributed for all chiropractic practices in Maryland, estimate a 95% confidence interval for the mean amount charged by Maryland chiropractors (t (15, 0.05) = 2.13):