suppose that the mean weight of king penguins found in an anarctic colony last year was 15.4 kilograms and the standard deviation was 2.5 kilograms. you want to test the null hypothesis that the mean penguin weight this year does not differ from last year. assume that the standard deviation has not changed. you are planning to measure a random sample of 100 penguins. at 10% significance level, find the critical values for the test.

A popular deviation (or σ) is a degree of the way dispersed the facts is on the subject of the imply. Low popular deviation way facts are clustered across the imply, and excessive popular deviation shows facts are greater unfold out. What is the usual deviation example? Consider the facts set: 2, 1, 3, 2, four.

The imply and the sum of squares of deviations of the observations from the imply could be 2.four and 5.2, respectively. Thus, the usual deviation could be √(5.2/5) = 1.01. Standard deviation tells you ways unfold out the facts is. It is a degree of the way some distance every determined fee is from the imply. In any distribution, approximately 95% of values could be inside 2 popular deviations of the imply.

Given sample mean=x=15.4 kg

sample standard deviation=s=2.5 kg

sample size=n=100 penguins

z multiplier for 90 CI is 1.645

90% CI for population mean is given by:

[tex]X+Z_{\frac{\alpha }{2} } * \frac{Z}{\sqrt[]{N} }[/tex]

15.4 (±)1.645(2.5)/sqrt(100)

15.4(±)1.645(2.5)/10

15.4 (±) 0.41125

15.4-.41125<μ<15.4+.41125

14.98875<μ<15.81125

we are 90% confident that the mean penguin height lies in the range

14.99 and 15.81

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