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)

19.

step-by-step explanation:

you add the 9 in the tens column since it is a single digit and 9 plus zero is 9 so it is 19

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct