The average bill in a restaurant for 49 customers is found to be $30.90, with a...
The average bill in a restaurant for 49 customers is found to be $30.90, with a standard deviation of $3.20. Construct a 99% confidence interval for the average bill.
Homework Answers
therefore 99% confidence
interval for average bill (29.7206, 30.0794)Add Answer to:
The average bill in a restaurant for 49 customers is found to be
$30.90, with a...
The mean bill of 16 randomly selected customers in a restaurant is found to be $32,...
The mean bill of 16 randomly selected customers in a restaurant is found to be $32, with a standard deviation of $3.50. Construct a 99% confidence interval for the mean bill of a customer.
A random sample of 49 lunch customers was taken at a restaurant. The average amount of...
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes. a. Compute the standard error of the mean. b. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant. c. With a .99 probability, how large of a sample would have to be taken to provide a...
A random sample of 13 customers in a Kentucky fast food restaurant revealed an average bill...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
A survey of 25 randomly selected customers found that their average age was 31.84 years with...
A survey of 25 randomly selected customers found that their average age was 31.84 years with a standard deviation of 9.84 years. What would the critical value t* be for a 95% confidence interval with 99 degrees of freedom? What would the margin of error be for a 95% confidence interval for this data (with the sample size of 25)? Construct a 95% confidence interval for the mean age of all customers for this data (with the sample size of...
The owner of a restaurant wants to determine the average number of customers who eat lunch...
The owner of a restaurant wants to determine the average number of customers who eat lunch at his restaurant each day. A sample of 16 days showed that an average of 130 lunches were served daily. The standard deviation of the sample was 12, Provide a 90% confidence interval estimate for the average number of lunches served per day. If the standard deviation for lifetimes of vacuum cleaners is estimated to be 400 hours, how large a sample must be...
A study compared the cost of restaurant meals when people pay individually versus splitting the bill...
A study compared the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they they would each be responsible for individual meals costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54, while the 24 people...
years. A survey of 25 randomly selected customers found the ages shown (in years). The mean...
years. A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.84 years and the standard deviation is 8.84 25 39 43 25 42p a) Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 23 28 42 45 31 interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed...
An Internet server conducted a survey of 400 of its customers and found that the average...
An Internet server conducted a survey of 400 of its customers and found that the average amount of time spent online was 12.5 hours per week with a standard deviation of 5.4 hours. a. Construct a 95% confidence interval for the average online time for all users of the particular Internet server. b. If the Internet server claimed that its users averaged 15 hours of use per week, would you agree or disagree? Explain.
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses...
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses in the amount of € 43.50 with a standard deviation of 19.20. a) Assuming that the dining expenses are normally distributed, construct a 95% confidence interval for the population mean expenses among xyz consumers( keep 2 decimals in your answer) b) How confident are you that the estimated 43.50 will be within 1.50 of the population mean? c) What would your margin of error...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 33.60 years and the standard deviation is 9.52 years a) Construct a 90% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 years?...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
The owner of a restaurant wants to determine the average number of customers who eat lunch at his restaurant each day. A sample of 16 days showed that an average of 130 lunches were served daily. The standard deviation of the sample was 12, Provide a 90% confidence interval estimate for the average number of lunches served per day. If the standard deviation for lifetimes of vacuum cleaners is estimated to be 400 hours, how large a sample must be...
years. A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.84 years and the standard deviation is 8.84 25 39 43 25 42p a) Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 23 28 42 45 31 interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed...
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses in the amount of € 43.50 with a standard deviation of 19.20. a) Assuming that the dining expenses are normally distributed, construct a 95% confidence interval for the population mean expenses among xyz consumers( keep 2 decimals in your answer) b) How confident are you that the estimated 43.50 will be within 1.50 of the population mean? c) What would your margin of error...
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 33.60 years and the standard deviation is 9.52 years a) Construct a 90% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 years?...
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