What is the z multiplier for a 70% confidence level? Round to the nearest 3rd decimal place, x.xxx
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- A poll for the presidential campaign sampled 491 potential voters in June. A primarypurpose of the poll was to obtain an estimate of the proportion of potential voters whofavored each candidate. Assume a planning value of p* = .50 and a 95% confidencelevel.a. For p* = .50, what was the planned margin of error for the June poll?Audio cassette tapes of a particular brand are claimed by the manufacturer to give, onaverage, at least 60 minutes of playing time. After receiving some complaints, themanufacturer’s quality control manager obtains a random sample of 64 tapes and measuresthe playing time, ? minutes, of each. The results are summarized by∑ ? = 3953.28 and ∑ ?² = 244,557.00(a) Construct a symmetric 99% confidence interval for the population mean playing time ofthe brand of tape. 2(b) Does the confidence interval support the customer’s complaints? Give a reason for youranswer.1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are deliveredon time by the Postal Service.
- Out of a 54 person sample from our fall Math-150 classes, 12.4% of the students are left-handed. Using a TI 84 graphing calculator: Construct an 80% confidence interval for the population proportion, p. Round to tenths of a percentand write a sentence answer.The U.S. has 13.1% of it’s population that are left-handed. Did this fall in your confidence interval from part a? Be sure to show all calculator commans used.Given a large sample with a mean of 72 and a SEM of 10, input the smallest whole number (no decimal) that would fall outside a 95% confidence interval? It can either the upperbound or lowerbound.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…
- The National Weight Control Registry tries to mine secrets of successfrom people who lost at least 30 pounds and kept it off for at least a year.It reports that out of 2,720 registrants, 439 were on a low-carbohydrate diet (less than 90 grams a day). a. Develop a 95% confidence interval for the proportion ofpeople on a low-carbohydrate diet. b. Is it possible that the population percentage is 16%? c. How large a sample is needed to estimate the proportionwithin 0.55%?Mrs. Voorhees claims that a Bowie knife is the most effective method of disposing rowdy campers. In a random sample of campers at Lake Crystal, a statistician found that a 99% confidence interval for the proportion of campers disposed by a Bowie knife is 0.7326 to 0.8154. Which of the following is the correct interpretation for this confidence interval? A. We are 99% confident that the population proportion of campers disposed by a Bowie knife will be within the interval. B. 99% of the time, the sample proportion of campers disposed by a Bowie knife will be in the interval. C. There is a 99% probability that the population proportion of campers disposed by a Bowie knife will be within the interval. D. We are 99% confident that the sample proportion of campers disposed by a Bowie knife will be within the interval.in recent poll of 900 randomly selected high school athletes, 86% said they regularly workout on the weekend. what is the margin of error. using a 95% confidence level, for estimating the true population proportion of athletes who regularly workout on the weekends? A. 0.023 B. 0.033 C. 0.011 d. 0.014 2. a researcher carried out a hypothesis test using two tailed alternatives hypothesis. which of the following z- scores is associated with the largest p- value? a. z=-0.01 B. z=2.71 c. z=-2.45 d. z= 0.05
- If n = 340 and ˆpp^ (p-hat) = 0.6, construct a 95% confidence interval. According to the Oxnard College Student Success Committee report in the previous year, we believe that 21% of students struggle in their classes because they don't spend more than 8 hours studying independently outside of a 4-unit class. For this year, you would like to obtain a new sample to estimate the proportiton of all Oxnard students who struggle in their classes because they don't study enough outside of the classrooms. You would like to be 99% confident that your estimate is within 5% of the true population proportion. How large of a sample size is required? Do not round mid-calculation.Platinum Gym has 10,000 gym members out of which 1500 memberships included Unlimited Fitness Training and use of the tanning salon, and out of which 750 included Unlimited Hydromassage. If the Fitness Training is considered A, the use of the tanning salon is considered B, and the Hydromassage is considered C, then the associate rule for these sales becomes "If A and B are purchased, then C is also purchased." Calculate the confidence level. A. 0.2 B. 0.075 C. 0.15 D. 0.5In a recent survey of US adults, 50 people were asked the question: “Do you use a ride sharing app?”. Of those surveyed, 44 responded that they did use a ride sharing app. Use the “plus-four” method to find a 99.7% confidence interval for the true proportion of US adults who use a ride sharing app. Round to three decimal places, if necessary. z0.10 z0.05 z0.025 z0.01 z0.005 z0.003 z0.0015 1.282 1.645 1.960 2.326 2.576 2.748 2.968 Hint: Remember the Empirical Rule, which says that if the distribution is unimodal and symmetric, then: Approximately 68% of the observations (roughly two-thirds) will be within one standard deviation of the mean. Approximately 95% of the observations will be within two standard deviations of the mean. Nearly all the observations will be within three standard deviations of the mean.
