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A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are members of
A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 276 members were looked at and their mean number of visits per week was 3.4 and the standard deviation was 2.8. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a distribution. b. With 9695' confidence the population mean number of visits per week is between and visits. c. If many groups of 276 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of visits per week and about percent will not contain the true population mean number of visits per week. For a confidence Level of 30.13 with a sample size 0f19, find the critical t value. A psychiatrist is interested in finding a 90% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 13 randomly selected children with Tourette syndrome. Round answers to 3 decimal places where possible. "I\""Illllllll a. To compute the confidence interval use a distribution. b. 1W'ith 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between l and I l. c. If many groups of 13 randomly selected children with Tourette syndrome are observed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of tics per hour and about percent will not contain the true population mean number of tics per hour. A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 35 specimens and counts the number of seeds in each. Use her sample results [mean = 66.5, standard deviation = 18.5] to find the 805% confidence interval for the number of seeds for the species. Enter your answer as an open- interval {i.e., parentheses] accurate to 3 decimal places. 80?; C.|. =I
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