# Question

According to a survey by Runzheimer International, the average cost of a fast-food meal (quarter-pound cheese-burger, large fries, medium soft drink, excluding taxes) in Seattle is $4.82. Suppose this figure was based on a sample of 27 different establishments and the standard deviation was $0.37. Construct a 95% confidence interval for the population mean cost for all fast-food meals in Seattle. Assume the costs of a fast-food meal in Seattle are normally distributed. Using the interval as a guide, is it likely that the population mean is really $4.50? Why or why not?

## Answer to relevant Questions

A survey of 77 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.48 minutes. The population standard deviation was 12 minutes. Construct a 95% confidence interval for the ...A bank officer wants to determine the amount of the average total monthly deposits per customer at the bank. He believes an estimate of this average amount using a confidence interval is sufficient. How large a sample should ...A company has developed a new light bulb that seems to burn longer than most residential bulbs. To determine how long these bulbs burn, the company randomly selects a sample of these bulbs and burns them in the laboratory. ...a. Use the data given to test the following hypotheses.b. Use the p-value to obtain the results.c. Solve for the critical value required to reject themean.A random sample of 51 items is taken, with x̄ = 58.42 and s2 = 25.68. Use these data to test the following hypotheses, assuming you want to take only a 1% risk of committing a Type I error and that x is normally ...Post your question

0