Mathematics, 27.03.2020 00:28, tsunommygames
The mean amount of money spent on lunch per week for a sample of 592592 students is $15$ 15. If the margin of error for the population mean with a 98%98% confidence interval is 1.701.70, construct a 98%98% confidence interval for the mean amount of money spent on lunch per week for all students.
Answers: 3
is the mean of the sample and M is the margin of error.
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The mean amount of money spent on lunch per week for a sample of 592592 students is $15$ 15. If the...
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