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Chapter 8: Q 79. (page 492)

Willows in Yellowstone Writers in some fields summarize data by giving x ̄ and its

standard error rather than x¯andSX.Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x¯andSX The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55±19.03. The researchers measured a total of 23plants.

a. Find the sample standard deviation sxfor these measurements.

b. A hasty reader believes that the interval given in the table is a 95% confidence interval for the mean height of willow plants in this region of the park. Find the actual confidence level for the given interval.

Short Answer

Expert verified

a. Standard deviation is 91.2547mm

b. Actual confidence level is 66%

Step by step solution

01

Given Information

It is given that:

SE=19.03

x¯=61.55

n=23

02

Standard Deviation

We know that SE=sxn

19.03=sx23

sx=19.03×23

sx=91.2647

03

Actual Confidence Level

The degree of freedom is =df=n-1=23-1=22

Using excel formula, confidence level is found as =2*TDIST(1,22,2)

Confidence Level is66%

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