Question

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.

Answer 1

Answer:

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

Step-by-step explanation:

We are given that

n=45

$\mu=1192$

Standard deviation,$\sigma=279$

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

$(\mu\pm Z\frac{\sigma}{\sqrt{n}})$

Using the formula

98% confidence interval is given by

$=(1192\pm 2.326\times \frac{279}{\sqrt{45}})$

$=(1192\pm 96.740)$

=$(1192-96.740,1192+96.740)$

=$(1095.260,1288.740)$

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