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Chapter 6: Q 4E (page 338)

A random sample of 90 observations produced a mean x = 25.9 and a standard deviation s = 2.7.

a. Find an approximate 95% confidence interval for the population meanμ

b. Find an approximate 90% confidence interval forμ

c. Find an approximate 99% confidence interval forμ

Short Answer

Expert verified

The standard deviation is commonly utilized as a measurement of an asset's comparative volatility. The standard deviation is determined as the square root of the variation by calculating the departure of every observation point from the mean.

Step by step solution

01

(a) The data is given below

Sample sizen=90X=25.9s=2.7

Here, the population standard deviation is known in this case, we use the t interval.

Degrees of freedom = n-1=90-1=89

The formula is given below:

X-E,X¯+E

Given the error in the margin

E=tc×sntc=1.986

E=1.986×2.790=5.36229.486833=0.565226

The confidence interval is: 25.9-0.565226, 25.9 + 0.565226=>25.3348,26.4652

02

(b) The data is given below

The formula is given below:

X-E,X¯+E

Given the error in the margin

E=tc×sntc=1.662

E=1.662×2.790=4.48749.486833=0.473013

The confidence interval is:25.9-0.473013, 25.9 + 0.473013= >25.4270,26.3730

03

(c) The data is given below

The formula is given below:

χ-E,χ+E

Given the error in the margin

E=tc×sntc=1.662E=2.632×2.790=7.10649.486833=0.74908

The confidence interval is: (25.1509, 26.6491)

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