Question

A random sample is selected from a normal popula-tion with a
mean of μ = 40 and a standard deviation of σ = 10. After a
treatment is administered to the individuals in the sample, the
sample mean is found to be M = 46.

How large a sample is necessary for this sample mean to be
statistically significant? Assume a two-tailed test with alpha =
.05.

Answer #1

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The normal distribution parameters are given as:

Mean, Mu = 40

Stdev, Sigma = 10

Also given

Xbar, M = 46

n = ?

alpha = .05

We will these parameters along with the standardization formula to solve the problem. The formula for standardization is : Z = (X-Mean)/(Stdev/sqrt(n))

The Z for .05 ( 2tailed) is -1.96

So, test-statistic should be more than 1.96 in magnitude

So, test-statistic =

(M-Mu)/(Sigma/sqrt(n))

= (46-40)/(10/sqrt(n)) > 1.96

Squaring and juggling n we get :

0.6*sqrt(n) > 1.96

n> (1.96/.6)^2 = 10.67 or n = 11

**Answer: The sample size should be atleast 11 in size for
the sample mean to statistically significant**

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