Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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.

Sagot :

Answer:

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

Step-by-step explanation:

We are given that

n=45

[tex]\mu=1192[/tex]

Standard deviation,[tex]\sigma=279[/tex]

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

[tex](\mu\pm Z\frac{\sigma}{\sqrt{n}})[/tex]

Using the formula

98% confidence interval is given by

[tex]=(1192\pm 2.326\times \frac{279}{\sqrt{45}})[/tex]

[tex]=(1192\pm 96.740)[/tex]

=[tex](1192-96.740,1192+96.740)[/tex]

=[tex](1095.260,1288.740)[/tex]

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

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.