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The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years. Based on the historic data it is known that the population standard deviation is 5 years. The margin of error of a 98 percent CI for the true mean client age is approximately:

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Answer:

2.33

Step-by-step explanation:

We shall use the z distribution because we know the population standard deviation.

The margin of error formula is given by:

[tex]E=Z_{\frac{\alpha}{2}}(\frac{\sigma}{\sqrt{n}})[/tex]

The z-value for 98% confidence interval is 2.326.

The sample size is n=25.

The population standard deviation is is

[tex] \sigma = 5[/tex]

We substitute the values into the formula to get:

[tex]E=2.326(\frac{5}{\sqrt{25}})[/tex]

Evaluate square root:

[tex]E=2.326(\frac{5}{5})[/tex]

Simplify:

[tex]E=2.326(1) = 2.326[/tex]

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