A company wishes to estimate the average amount the customers pay per month. A random sample of 200 customers' bills produced a sample mean of $50. Suppose that the population standard deviation is known and equals $5. a) Find a 95% confidence interval for the population mean payment. A company wishes to estimate the average amount customers pay per month. A random sample of 200 customers' bills produced a sample mean of $50 and a sample standard deviation of $5. b) Without any calculations, do you expect that a 95% confidence interval for the population mean will be wider than in the previous question, or narrower, or the same? A company wishes to estimate the average amount customers pay per month. A random sample of 100 customers' bills produced a sample mean of $50 and a sample standard deviation of $5. c) Find a 95% confidence interval for the population mean.9 Take t*=1.980. The answers below may present approximate values

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 14PPS
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A company wishes to estimate the average amount the customers pay per month. A random sample of 200 customers' bills  produced a sample mean of $50. 

Suppose that the population standard deviation is known and equals $5. 

a)  Find a 95% confidence interval for the population mean payment. 

A company wishes to estimate the average amount customers pay per month. A random sample of 200 customers' bills  produced a sample mean of $50 and a sample standard deviation of $5. 

 b)  Without any calculations, do you expect that a 95% confidence interval for the population mean will be wider than in the previous question, or narrower, or the same? 

A company wishes to estimate the average amount customers pay per month. A random sample of 100 customers' bills  produced a sample mean of $50 and a sample standard deviation of $5. 

 c)  Find a 95% confidence interval for the population mean.9 Take t*=1.980. The answers below may present approximate values. 

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