A fitness center is interested in finding a 98% 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 252 members were looked at and their mean number of visits per week was 2.5 and the standard deviation was 1.9. a. To compute the confidence interval use a ? distribution. b. With 98% confidence the population mean number of visits per week is between and visits. c. If many groups of 252 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 98% 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 252 members were looked at and their mean number of visits per week was 2.5 and the standard deviation was 1.9. a. To compute the confidence interval use a ? distribution. b. With 98% confidence the population mean number of visits per week is between and visits. c. If many groups of 252 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.
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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![A fitness center is interested in finding a 98% 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 252 members were looked at, and their mean number of visits per week was 2.5, with a standard deviation of 1.9.
a. To compute the confidence interval, use a t-distribution.
b. With 98% confidence, the population mean number of visits per week is between [ ] and [ ] visits.
c. If many groups of 252 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7248d714-2630-4255-aa39-67ac8596a103%2Fc612bac7-24a0-47a3-9bb9-519dbce3bf01%2F9t9k5m8_processed.png&w=3840&q=75)
Transcribed Image Text:A fitness center is interested in finding a 98% 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 252 members were looked at, and their mean number of visits per week was 2.5, with a standard deviation of 1.9.
a. To compute the confidence interval, use a t-distribution.
b. With 98% confidence, the population mean number of visits per week is between [ ] and [ ] visits.
c. If many groups of 252 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.
![**Problem Statement:**
You measure 36 watermelons' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 11.2 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.
Give your answer as a decimal, to two places.
± ___ ounces
---
This problem involves calculating the margin of error for a confidence interval given certain statistical information. Specifically, we are looking for the maximal margin of error for a 90% confidence interval. To solve this, use the formula for the margin of error:
\[ \text{Margin of Error} = Z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( Z \) is the Z-score corresponding to the desired confidence level (for 90% confidence, \( Z \approx 1.645 \)).
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
**Given:**
- Mean weight = 79 ounces
- Population standard deviation (\( \sigma \)) = 11.2 ounces
- Sample size (\( n \)) = 36
- Confidence level = 90%
Use this information to calculate the margin of error and fill in the blank provided in the diagram.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7248d714-2630-4255-aa39-67ac8596a103%2Fc612bac7-24a0-47a3-9bb9-519dbce3bf01%2Fdaox5tc_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
You measure 36 watermelons' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 11.2 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.
Give your answer as a decimal, to two places.
± ___ ounces
---
This problem involves calculating the margin of error for a confidence interval given certain statistical information. Specifically, we are looking for the maximal margin of error for a 90% confidence interval. To solve this, use the formula for the margin of error:
\[ \text{Margin of Error} = Z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( Z \) is the Z-score corresponding to the desired confidence level (for 90% confidence, \( Z \approx 1.645 \)).
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
**Given:**
- Mean weight = 79 ounces
- Population standard deviation (\( \sigma \)) = 11.2 ounces
- Sample size (\( n \)) = 36
- Confidence level = 90%
Use this information to calculate the margin of error and fill in the blank provided in the diagram.
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