Question

In: Math

A geneticist interested in human populations has been studying the growth patterns in American males since...

A geneticist interested in human populations has been studying the growth patterns in American males since 1900. A monograph
written in 1902 states that the mean height of adult American males is 67.0 inches with a standard deviation of 3.5 inches. Wishing
to see if these values have changed over the 20th century, the geneticist measured a random sample of 28 adult American males
and found that the sample mean was 69.4 inches and the sample standard deviation was 4.0 inches.
Considering the 1902 data to be a population, do the more recent data suggest that the height of American males has significantly
changed?
Put your answers in column B
Ho:
Ha:
test-statistic*:
df:
Exact P value for the test-statistic:
Conclusion relative to the hypothesis:
(Don't forget your parenthetical ts= ,df= ,P=
support statement)
*test-statistic refers to the statistical test value for whatever statistical test is done to answer the question.
What is the Statistical Power of this test?: %

Solutions

Expert Solution

We test whether the population mean has changed or not.

Thus considering 1902 data as a population we get,

Null hypothesis:

Alternative hypothesis:

Test statistic:

We have,

n = 28, , s = 4

Plugging in values we get,

Degrees of freedom:

df = n - 1 = 28 - 1 = 27

p-value:

Consider 5% level of significance:

0.0038 < 0.05

i.e.

Hence we reject null hypothesis.

There is a sufficient evidence to conclude that "height of American males has significantly
changed".

Power of the test:

Let us assume that is true. i.e.

Corresponding z for 0.05 level of significance: (for two tailed test)

Let us assume 'b' such that,

and

Assume H0 is false and instead

Hence power of the test is probability that b < 65.703 or b > 68.30

And

Power of the test is the probability that z is greater than -1.66 and z is less than -5.59

We know that P(z < -5.59) = 0

Therefore,

Hence power of the test is 95.15%


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