The expected mean of a normal population is 100, and its standard deviation is 1

Question

The expected mean of a normal population is 100, and its standard deviation is 10. A sample of 25 measurements gives a sample mean of 96.

How many measurements need to be mean so that the .05 level test of Ho (null hypo) Uo=100 has a 80% probability of rejection when U is really 97?

I believe it involves a using the second type error to solve for Beta, and you solve for this in the previous question that says
What is the probability of type II error when: U is really 97 and level of signifiance .05.

Thanks

Explanation / Answer

Given that,
Standard deviation, =10
Sample Mean, X =96
Null, H0: <100
Alternate, H1: >100
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-100)/10/(n) < -1.6449 OR if (x-100)/10/(n) > 1.6449
Reject Ho if x < 100-16.449/(n) OR if x > 100-16.449/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 25 then the critical region
becomes,
Reject Ho if x < 100-16.449/(25) OR if x > 100+16.449/(25)
Reject Ho if x < 96.71 OR if x > 103.29
Implies, don't reject Ho if 96.71 x 103.29
Suppose the true mean is 97
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(96.71 x 103.29 | 1 = 97)
= P(96.71-97/10/(25) x - / /n 103.29-97/10/(25)
= P(-0.145 Z 3.145 )
= P( Z 3.145) - P( Z -0.145)
= 0.9992 - 0.4424 [ Using Z Table ]
= 0.557
For n =25 the probability of Type II error is 0.557

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