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In: Statistics and Probability

The Department of Natural Resources in New York wants to estimate the average length of lake...

The Department of Natural Resources in New York wants to estimate the average length of lake trout in Keuka Lake. They randomly take a sample of 100 lake trout and find the sample average length to be 18.3 inches with a sample standard deviation of 2.5 inches. Calculate a 99 % confidence interval for the average length of lake trout in Keuka Lake.

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Expert Solution

c )solution

Given that,

= 18.3

s =2.5

n = 100

Degrees of freedom = df = n - 1 =100- 1 = 99

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,99 =2.626    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.626 * ( 2.5/ 100) = 0.66

The 99% confidence interval is,

- E < < + E

18.3 -0.66 < < 18.3+0.66

(17.64,18.96)

orchestra answered 1 year ago
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