Point estimation of the mean

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A survey of American planning long summer vacations in 2002 revealed a mean planned expenditure of $1076.

Assume that this mean is based on a random sample of 300 Americans who were planning long summer vacations in 2002 and that the sample standard deviation was $345.

(a) What is the point estimate of the mean planned expenditure by all Americans taking long summer vacations in 2002?

(b) What is the 95% margin of error for the above mean planned expenditure?

(c) Construct a 99% confidence interval for the mean planned expenditure of 2002.

(d) What does the 99% confidence interval mean in this context?

(e) What is the maximum error of estimate for part (c)? Suppose the standard deviation of all Americans' planned expenditure of 2002 is believed to be $400 according to some experts.

(f) What, then, is the 99% confidence interval for the mean planned expenditure of 2002?

(g) What is the maximum error of estimate for part (f)?

Reference no: EM1386372

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