It was observed that 4 out of 10 randomly selected students went to the cinema. Find the 95% confidence interval estimate for the proportion of moviegoers in the population, \pi. P( 0,20 <1< 0,30 ) = 0,95 OA) P( 0,30
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- An SRS of 100 postal employees found that the average time these employees had worked for the postal service was x = 7 years with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately Normal. A 95% confidence interval for the mean time u the population of postal service employees have spent with the postal service is 7 + 0.525 7 + 0.4 7 + 2 7 + 1.984 bA sample of size 65 gave a mean 7.4 and standard deviation as 1.6 then 95% confidence interval for population mean will be Select one: O (17.008, 17.792) O [27.008, 30.008] O (7.008, 7.792) O (9,265, 10.735)In a data management exam, the scores of 567 participants followed a continuous normal distribution. It is known that the mean from all of the students is 75, and to reach the top 1%, students need to score more than (exactly) 92. a) Estimate the number of students who scored between 60 and 70. b) 10 students are good friends. After knowing each other’s marks, they came up with a 95% confidence interval of 65 ± 12 for the population mean. Can you conclude that, with a 90% confidence level, that these students performed worse than average?
- A research company has carried out a poll to find out about people’s economic outlook. 894 of 1854 young adults (aged between 20–29 years) said that the economic situation is likely to improve over the next year. By contrast, 1292 of 2243 mature adults (aged between 30–39 years) agreed with this statement.) Calculate the upper limit of an 80% confidence interval for the difference between the proportions of people holding this optimistic view in the two age groups in the population. Form the confidence interval around a positive point estimate for the difference between the proportions. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample proportions. Provide your answer as a number between 0 and 1 rounded to 4 decimal places.A survey found that out of a random sample of 164 workers, 141 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the upper bound limit of the 95% confidence interval of the population proportion of workers who are interrupted three or more times an hour. Round your answer to 3 decimal places.A random sample of size 30 from a normal population yields = 39 and s = 4.9. The lower bound of a 95 percent confidence interval is (Round off upto 2 decimal places).
- A random sample of 100 accounts from a branch of certain bank, has mean balance of Rs. 18,600 with S.D.= 1160. A random sample of 80 accounts from another branch of same bank has mean balance of Rs. 15,200; with S.D.= Rs. 920. Find 90% confidence interval of difference between Mean balance of all accounts in both branches.Express the confidence interval 0.84 ≠ 0.085 in open interval form (i.e., (0.155,0.855)).Let X1,., X, be iid from Normal(e,0) population (0>0). Use the pivotal quantity to find (1-a) Confidence Interval for 0. ...
- a. What is the probability of seeing positive change? b. What is the probability of seeing positive change given that the tree is oak? c. Do the data suggest effect and tree type are independent events? yes or noA random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98 % confidence interval for the true average age of all students in the university is * 20.5 to 29.5. O24.4 to 25.6. 23.0 to 27.0. O 20.0 to 30.0.A simple random sample from a population with a nomal distribution of 100 body temperatures has x= 98.20°F and s 0.64°F. Construct a 98% confidence interval estimate of the standard deviation of body temperature of alI healthy humans.