Question
The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meal per week. Construct a $99 \%$ confidence interval for the population mean.
Step 1
The sample mean ($\bar{x}$) is 2.76 meals per week, the sample standard deviation (s) is 0.75 meal per week, the sample size (n) is 60 couples, and the confidence level is 99%. Show more…
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The American Restaurant Association collected information on the number of meals eater outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meal per week. Construct a 99% confidence interval for the population mean.
The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meals per week. Construct a 97 percent confidence interval for the population mean.
Based on a survey of 1000 adults by Greenfield Online and reported in a May 2009 USA Today Snapshot, adults 24 years of age and under spend a weekly average of $35$dollar on fast food. If 200 of the 1000 adults surveyed who were in the 24 and under age category provided a standard deviation of $14.50$dollar, construct a $95 \%$ confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under. Assume fast food weekly expenditures are normally distributed.
Inferences Involving One Population
Inferences about the Mean $\mu(\sigma$ Unknown)
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