Use th e confidence level and sample data to find a confidence interval for estimating the...
Use the confidence level and sample data to find a confidence interval for estimating the population . Round your rower to the same number of decimal place as the sample mean From packages received by a percel service, 41 are randomly selected. The sample has a mean weight of 11.6 pounds. The population standard deviation is a 24 pounds. What is the 96% confidence interval for the tree w packages received by the parol service? O A 1075*2125 OB. 10.95<<123...
Use the confidence level and sample data to find a confidence interval for estimating the population u. Round your answer to the same number of decimal places as the sample mean. 20) A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 20) 5.2 on a placement test. What is the 90% confidence interval for the mean score,, of all students taking the test? A) 27.8 <u <31. 2 B ) 28.2<u<30....
Question 1 (a) In a metal fabrication process, metal rods are produced that have an average length of 20.5 feet with a standard deviation of 2.3 feet. A quality control specialist collects a random sample of 30 rods and measures their lengths. (i) Describe the sampling distribution for the sample mean by naming the model and telling its mean and standard deviation. [3 marks] (ii) Calculate the probability that the sample mean length of metal rods is less than 19.5...
7 points Save use entonce leve, and sample data to find a oorhdence inten-for estin ating population μ Rond your answer re randomly selected from packagens rcerved by a parcel service The samle has a mean weight of 29 9 pounds and a starndard deviation of 3.0 pounds weight, u, of all packages received by the parcel 307 288b<p <31.0
QUESTION 8 Use the confidence level and sample data to find a confidence interval for estimating the population u. Round your answer to the same numbe of decimal places as the sample mean. A random sample of 1 17 full grown lobsters had a mean weight of 22 ounces and a standard deviation of 2.7 ounces. Construct a 98% confidence interval for the population mean . ?20 oz < ? < 22 oz 022 oz < ? < 24 oz...
Use the confidence level and sample data to find a confidence interval for estimating the population. Round your answer to the same number of decimal places as the sample mean 19) A group of 5 randomly selected students have a mean score of 295 with a standard deviation of 19 5.2 on a placement test. What is the 90% confidence interval for the mean score of all students taking the test? A) 27.8 < <312 ) 27.9<<311 C)282 << 308...
Use the confidence level and sample data to find a confidence interval for estimating the population. Round your answer to the same number of decimal places as the sample mean A random sample of 87 light bulbs had a mean life of 479 hours with a standard deviation of a = 30 hours. Construct a 90% confidence interval for the mean life, ja, of all light bulbs of this type O A 472 hr < < 486 he B. 474...
Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104, 7 = 95.3, 0 = 6.5; 99% confidence 93.7 < < 96.9 O94.1 << 96.5 93.8 < p < 96.8 94.2 << 96.4
Question 12 8 pts A study of 80 mice showed that their average weight was 6.8 ounces. The standard deviation of the population is 0.9 ounces. Which of the following is the 99% confidence interval for the mean weight per mouse? (5.4408, 8.7502) (6.0408,7.9502) (6.5408, 7.0592) (5.9408,8.1502) Question 13 8 pts 57 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 13,5 pounds; the population standard deviation for all packages is...
1. Use the given degree of confidence and sample data to construct a confidence interval for the point) population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. 0 0.438<p0.505 0 0.444 p0.500 0 0.435<p<0.508 O 0.471 p0.472 2. Use the given data to find the minimum sample size required...