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kramer
3 years ago
5

Suppose a research company takes a random sample of 45 business travelers in the financial industry and determines that the samp

le 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.
Mathematics
1 answer:
Rufina [12.5K]3 years ago
7 0

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)

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