The mean amount of money spent on lunch per week for a sample of 75 students is $42. If the margin of error for the population mean with a 99% confidence interval is 2.10, construct a 99% confidence interval for the mean amount of money spent on lunch per week for all students.
Confidence interval formula is
where , E= margin of error = 2.10
Substitute here known values
=>(39.9,44.1)
Therefore, a 99% confidence interval for the mean amount of money spent on lunch per week for all students
is ($39.9 , $44.1)
The mean amount of money spent on lunch per week for a sample of 75 students...
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