# I need help writing the confidence interval?A.) The assumed standard deviation = 0.5 Variable N Mean StDev SE Mean 90% CI Price 16 5.516 1.176 ...

I need help writing the confidence interval?

A.) The assumed standard deviation = 0.5

Variable N Mean StDev SE Mean 90% CI

Price 16 5.516 1.176 0.125 (5.310, 5.721)

Test of mu = 5 vs > 5

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### 1 Answer

We are given `mu=5,sigma=.5,bar(x)=5.516,n=16` . We want to test at the 90% confidence level `(alpha=.1)` whether the true mean is greater than 5.

(1) `H_0:mu=5` `H_1:mu>5`

(2) The critical point is `z_alpha=z_.1=1.28` Thus the critical region is `z>1.28`

(3) The test value is `z=(bar(x)-mu)/(sigma/sqrt(n))=(5.516-5)/(.5/sqrt(16))=4.128`

(4) The test value is in the critical region so we reject the null hypothesis

(5) There is sufficient evidence to conclude with a 90% certainty that `mu>5`

** If you meant for the standard deviation to be 1.176 then the critical region stays the same and the test value is `(5.516-5)/(1.176/sqrt(16))=1.755` which is still in the critical region.**

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If you really just want the confidence interval (no hypothesis test) then you have:

`bar(x)-z_(alpha/2)(sigma/sqrt(n))<=mu<=bar(x)+z_(alpha/2)(sigma/sqrt(n))` or

`5.516-1.64(.5/4)<=mu<=5.516+1.64(.5/4)`

`5.311<=mu<=5.721`

(Note the use of `alpha/2` )

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