Express the confidence interval (84.4 %, 95.2 % ) in the form of p + ME.
Q: A random sample of 40 students taken from a university showed that their mean GPA is 2.94 and the…
A: Obtain the 99% confidence interval for the population mean GPA of students: Denote the population…
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Q: Express the confidence interval 14.4<p<28.4% in the form of p+-ME %+- %
A: Point estimate p-hat: The formula for point estimate is,
Q: Express the confidence interval (68 % , 79.6 % ) in the form of p + ME.
A: The mid-point of the confidence interval gives the value of p-cap.
Q: A random sample of 300 high school students in a particular town showed that 165 smoke on a regular…
A: Given Information: Sample size (n) = 300 Proportion of students who smoke (p) = 165/300 = 0.55
Q: A random survey of 70 male sophomores showed 16 preferred off campus living and out of 63 female…
A:
Q: Express the confidence interval (30.6 % , 43.6 % ) in the form of p ± ME. % + %
A: Express the confidence interval (30.6% , 43.6%) in the form of p^±ME.
Q: In a sample of 270 adults, 151 had children. Construct a 90% confidence interval for the true…
A: Given that Sample size n =270 Favorable cases x =151 Sample proportion p^=x/n =151/270 =0.5593
Q: A 90% confidence interval for the mean proportion of airline reservations being canceled on the day…
A: Solution: From the given information, a 90% confidence interval for the mean proportion of airline…
Q: Express the confidence interval (50.6 %, 56.6 %) in the form of p + E. % +
A:
Q: In a survey, 22% of 230 single women said that they "definitely want to have children."In the…
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Q: A recent study of 2700 children randomly selected found 21% of them deficient in vitamin D. …
A: Given that, 2700 children randomly selected found 21% of them deficient in vitamin D. n=2700p^=0.21…
Q: i need help with the upper 95% confidence interval of the difference
A: We have given the sample mean of difference of scores X̅=-1.60 with standard deviations s=1.1402…
Q: Of 400 employees selected randomly from one company, 18.25% of them commute by carpooling. Construct…
A: The sample size n = 400. The probability p is 0.1825 The absolute value of Zα/2 is 1.96 using the…
Q: If n = 160 and p (p-hat) = 0.85, construct a 95% confidence interval. Give your answers to three…
A:
Q: If n = 320 and p (p-hat) = 0.7, construct a 90% confidence interval. Give your answers to three…
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Q: Determine from a confidence interval whether a test is significant?
A: Null hypothesis states that there is no difference between populations or a set of given…
Q: In a sample of 290 adults, 191 had children. Construct a 95% confidence interval for the true…
A: 0.604 < p < 0.713 Explanation:Approach to solving the question:To construct a 95% confidence…
Q: Express the confidence interval (74.6%,82.4%)in the form of ˆp±ME
A:
Q: In a sample of 550 adults, 512 had children. Construct a 99% confidence interval for the true…
A: Given X=512 n=550
Q: In a random sample of 500 Americans, 50% indicated that they have a cat for a pet. Estimate with 95%…
A: Given information
Q: FIND Za, FOR THE 94% /2 CONFIDENCE INTERVAL
A: Step 1: Subtract the confidence level from 100% to find the α level: 100% – 94% = 6%. Step 2:…
Q: press the confidence interval (81.8%,96.4%)(81.8%,96.4%) in the form of ˆp±E.
A: Given: Lower limit = 81.8% = 0.818 Upper limit = 96.4% = 0.964 The point estimate can be calculated…
Q: how would i do 19 b with a 95% confidence interval***** not 90%
A: Given Information B) Sample Mean(M) = 77.2 Sample Standard Deviation = 5.7 Sample size(n) = 9…
Q: An internet service provider sampled 1000 of its customers, 189 of them experienced an interruption…
A: It is given that sample size (n) is 1000 and x is 189.
Q: As the sample standard deviation increases when all other parts of the confidence interval stay the…
A: We want tell the sample standard deviation increases when all other parts of the confidence interval…
Q: Express the confidence interval (75.7 % , 86.5 % ) in the form of p + E. %土
A:
Q: Suppose that a garage finds that 42 customers of 142 customers samples get their oil changes on…
A: The objective of this question is to find the lower bound of the 98% confidence interval for the…
Q: A motorcycle dealer with the "Hd" brand wants to know the percentage of students in the city "X" who…
A: We have to find confidence interval.
Q: Express the confidence interval ( 26.2%, 35.2%) in the form of p-hat plus or minus M.E.
A: Given Conference interval = (26.2℅ , 35.2℅) =(0.262 , 0.352)
Q: A firm randomly samples 100 workers at its New York plant and finds that 32% of the workers say that…
A: The proportion of workers who say that they buy the firm’s product is 0.32.
Q: If a sample has has a 94% confidence interval of (4.68,5.43), what is the 2-decimal margin of error?
A: Given information- We have given the confidence interval is- C.I = (4.68, 5.43) Lower limit = 4.68…
Q: Suppose that a garage finds that 42 customers of 142 customers samples get their oil changes on…
A: The objective of this question is to find the lower bound of the 98% confidence interval for the…
Q: A survey found that 611 out of 663 college graduates cannot correctly spell the word aluminum. Find…
A: Given that A survey found that 611 out of 663 college graduates cannot correctly spell the word…
Q: Suppose that a garage finds that 30 customers of 138 customers samples get their oil changes on…
A: The objective of this question is to find the upper bound of the 98% confidence interval for the…
Q: A recent study by the McGraw-Hill company showed that 61% (1621) of college students use their…
A: Given, Proportion of college students use their smartphones for study = 0.61Total number of students…
Q: Does this interval give the quality control inspector reason to believe that the mean amount of…
A: The 95% confidence interval for the true mean amount of juice dispensed per bottle is (296.4,…
Q: During a period of 11 years, of 1700 people that were selected for grand jury duty, 30% of them were…
A: For large sample size population proportion will follow normal dustribution as per central limit…
Q: The mileage a car got on 5 trips from Greenwood to Greenville was used to create a 90% confidence…
A: Solution Given that 90% confidence interval is (32.32, 38.32).
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- In a sample of 460 adults, 281 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places < p <A random sample of 120 rhesus monkeys found that the number of rhesus monkeys that live in a captive breeding facility is 77. Construct a 90% confidence interval for the true proportion of rhesus monkeys that live in a captive breeding facility.FIND Za/2 FOR THE 86 % CONFIDENCE INTERVAL
- 2If n = 380 and ˆ = 0.25, construct a 95% confidence interval.< p < (Give your answers to three decimals)Researchers are studying whether the proportion of homeschooled children in district 1 (considered 1) is less than the proportion of homeschooled children in neighboring district 2 (considered as 2). To test the theory info was gathered: district 1 had 600 students of which 90 were homeschooled. District 2 had 550 students of which 99 were homeschooled. What is the 95% confidence interval for the difference proportion between district 1 and 2? A. (-0.086, 0.026) B. (0.143, 0.217) C. (0.113, 0.187)
- We randomly sample 600 people in a specific town. 342 of them like cats. Based on this, construct a 95% confidence interval for the true population proportion of people in this town who like cats. Give your answers as decimals, rounded to four places.Suppose that a garage finds that 30 customers of 138 customers samples get their oil changes on time. Find a 98% confidence interval of the true proportion. What is the upper bound of the confidence interval? (round to 4 decimal places)During the 2016 season, Lauren Robinson had 92 at bats and made 36 hits. Calculate a 95% confidence interval for her ABILITY to get hits in that season.
- In a sample of 360 adults, 266 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesThe general form of a confidence interval is inferred asIn a sample of 510 adults, 281 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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