Question

According to Runzheimer International, the average cost of a domestic trip for business travelers in the financial industry is $1,250. Suppose another travel industry research company takes a random sample of 51 business travelers in the financial industry and determines that the sample average cost of a domestic trip is $1,192, with a sample standard deviation of $279. Construct a 98% confidence interval for the population mean from these sample data. Assume that the data are normally distributed in the population. Now go back and examine the $1,250 figure published by Runzheimer International. Does it fall into the confidence interval computed from the sample data? What does this tell you?

(Round your answers to 2 decimal places.)

The 98% confidence interval:

≤ μ ≤

The figure given by Runzheimer International falls (within/outside) the confidence interval. Therefore, there is no reason to reject the Runzheimer figure as different from what we are getting based on this sample.

Answer 1

98% Confidence Interval is
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.02 /2, 51- 1 ) = 2.403
1192 ± t(0.02/2, 51 -1) * 279/√(51)
Lower Limit = 1192 - t(0.02/2, 51 -1) 279/√(51)
Lower Limit = 1098.12
Upper Limit = 1192 + t(0.02/2, 51 -1) 279/√(51)
Upper Limit = 1285.88

1098.12 < < 1285.88

The figure given by Runzheimer International falls within the confidence interval.

Therefore, there is no reason to reject the Runzheimer figure as different from what we are getting based on this sample.

answered by: anonymous