Suppose a research company takes a random sample of 45 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 (for domestic trip) from these sample data. Round your answers to 3 decimal places.

98% confidence interval for the population mean =(1095.260,1288.740)

Step-by-step explanation:

We are given that

n=45

Standard deviation,

We have to construct a 98% confidence interval for the population mean.

Critical value of z at 98% confidence, Z =2.326

Confidence interval is given by

Using the formula

98% confidence interval is given by

=

=

Hence, 98% confidence interval for the population mean (1095.260,1288.740)

i got 10.28 but that rounds up to 10.3

i hope this is what you were looking for.

see attached graph