A fitness center is interested in finding a 90% confidence interval for the mean number ofdays per week that Americans who are members of a fitness club go to their fitness center.Records of 265 members were looked at and their mean number of visits per week was 3.5and the standard deviation was 2.8. Round answers to 3 decimal places where possible.? distribution.a. To compute the confidence interval use ab. With 90% confidence the population mean number of visits per week is betweenandvisits.c. If many groups of 265 randomly selected members are studied, then a different confidenceinterval would be produced from each group. Aboutpercent of these confidenceintervals will contain the true population mean number of visits per week and aboutpercent will not contain the true population mean number of visits per week.

From the provided information, Sample size (n) = 265 Sample mean (x̄) = 3.5 Sample standard…

Answered: A fitness center is interested in…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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suppose you obtained a mean of $167 spent on transportation and a standard deviation of $40. Calculate a 95% confidence interval for the mean and select the values that come closest to those that would fill the spaces in the following interpretation: we can be 95% confident that they mean amount of money spent on transportation lies between _____ and _____ .
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 242 members were looked at and their mean number of visits per week was 3.4 and the standard deviation was 1.8. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a ? distribution. b. With 90% confidence the population mean number of visits per week is between visits. and c. If many groups of 242 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.
An irate patient complained that the cost of a doctor's visit was too high. He randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor's visit was $48.60. The standard deviation of the sample is $3.53. a. b. Find the best point estimate of the population mean. Construct and interpret the 85% confidence interval of the mean amount of money spent on a doctor's visit.
A random sample of 82 women showed that the mean number of children reported was 1.87 with a standard deviation of 1.7 . (Interestingly, a sample of 82 men showed a mean of 1.3 children.) Complete parts (a) through (c) below. LOADING... Click the icon to view the t-table. Question content area bottom Part 1 a. Find a 95% confidence interval for the population mean number of children for women. Because the sample size is so large, you can use 1.96 for the critical value of t (which is the same as the critical value of z) if you do the calculations manually. Choose the correct answer below and, if necessary, fill in the answer boxes to complete your choice. (Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.)
mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater systems has a mean price of $138.00. Assume the population standard deviation is $17.80. Construct a 90% confidence interval for the population mean.
Use the data attached: 1. In two to three sentences, explain what a confidence interval means(in general), as if you were talking to someone who has not taken statistics. 2. In one to two sentences, explain what this confidence interval means for this particular study.
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
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 268 members were looked at and their mean number of visits per week was 2.1 and the standard deviation was 1.8. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a ? 9 distribution. b. With 90% confidence the population mean number of visits per week is between and visits. c. If many groups of 268 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.
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 249 members were looked at and their mean number of visits per week was 2.2 and the standard deviation was 2.7. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a ? distribution. b. With 90% confidence the population mean number of visits per week is between and visits. c. If many groups of 249 randomly selected members are studied, then a different confidence percent of these confidence interval would be produced from each group. About 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.
In a survey, 28 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $45 and standard deviation of $14. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place. ________+________
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In a survey, 21 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $30.9 and standard deviation of $12.7. Estimate how much a typical parent would spend on their child's birthday gift (use a 90% confidence level). Give your answers to 3 decimal places. Express your answer in the format of a + E. $ 30.9 V + $ 4.770

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