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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 :

lublana

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)

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