A statistician chooses 27 randomly selected dates, and when examining the occupancy records of a particular motel for those dates, finds a variance of 34.34 and a standard deviation of 5.86 rooms rented. If the number of rooms rented is normally distributed, find the 95% confidence interval for the population standard deviation of the number of rooms rented. Interpret your interval in context.

The population variance of 95% confidence interval is and this can be determined by using the given data.

Given :

First, determine the degree of freedom.

The determine the confidence interval using the below formula:

Now, using the chi distribution the above expression becomes:

Simplify the above expression.

(21.297 ) ; (64.493)

So, the population variance of 95% confidence interval is:

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Confidence interval variance [21.297 ; 64.493]

Confidence interval standard deviation;

4.615, 8.031

Step-by-step explanation:

Given :

Variance, s² = 34.34

Standard deviation, s = 5.86

Sample size, n = 27

Degree of freedom, df = 27 - 1 = 26

Using the relation for the confidence interval :

[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]

From the chi distribition table :

X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844

Hence,

[34.34*26 / 41.923] ; [34.34*26 / 13.844]

[21.297 ; 64.493]

The 95% confidence interval for the population variance is :

21.297 < σ² < 64.493

Standard deviation is the square root of variance, hence,

The 95% confidence interval for the population standard deviation is :

4.615 < σ < 8.031

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step-by-step explanation:

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