Question

3. (15 pts) The salt content of a a sample standard deviation of s-48 (milligrams) (a) (5 pts) Test the hypothesis l 10: σ2-18; HI: ơ, 18 using. Pvalue ,.005 (b) (5 pts) S uppose the actual standard deviation is twice as large as the value listed in HO What is the power of the test? c) (5 pts) Suppose that the true variance is ơ2-40. How large would the samplesreooSo be in order to achieve a power of at least 0.9?

Answer 1

Solution

Back-up Theory

Power of a test = 1 – β = probability of rejecting a null hypothesis when it is not true, i.e., Alternative is true. Or equivalently, power is the probability of accepting a true alternative…………………. (1)

Now to work out the solution,

Given sample size, n = 10 sample stabdard deviation, s = 4.8 …………………………………… (2)

Part (a)

Let X = Ash content (in milligrams) of a prepared food.

Then, X ~ N(µ, σ²)

Claim: Population variance is 18

Hypotheses:

Null H₀: σ² = σ²₀ = 18 Vs Alternative H₁: σ² ≠ 18

Test statistic:

χ² = (n - 1)s²/σ²₀ where

n = sample size

s = sample standard deviation  

Calculations Summary

n =	10
σ20 =	18
s =	4.8
s^2 =	23.04
χ2(cal) =	11.52
Given
α =	0.05
Upper crit	19.02277
Lower crit	2.70039
α/2 =	0.025
1 - α/2 =	0.975
n - 1 =	9
p-value(U)	0.241741
p-value(L)	0.758259

Distribution, Significance Level, α, Critical Value and p-value

Under H₀, χ² ~ χ²_{n - 1}

Critical values = upper (α/2)% point and lower (α/2)% of χ²_{n - 1}.

p-value = P(χ²_{n – 1} > χ²_(cal)) or = P(χ²_{n – 1} < χ²_(cal))

Using Excel Function: Statistical CHIINV and CHIDIST, the above are found to be as shown in the above table.

Decision:

Since χ²crit(upper/lower) < χ²cal < χ²crit(upper/lower), or equivalently since p-value > α. H₀ is accepted.

Conclusion:

There is sufficient evidence to support the claim that the population variance is 18. Answer

Part (b)

Null hypothesis is rejected if {(n - 1)s²/σ²₀} > 19.0228 or {(n - 1)s²/σ²₀} < 2.700

Or, s² > 36.0456 or s² < 5.4 [just by substituting (n - 1) = 9 and σ²₀ = 18]

Now, under H₁ , σ², say σ²₁ = 36.

So, {(n - 1)s²/σ²₁} ~ χ²₉

Thus, s² > 36.0456 or s² < 5.4 is equivalent to saying: χ²₉ > 9.0114 or χ²₉ < 1.35

So vide (1),

Power = P(χ²₉ > 9.0114) + P(χ²₉ < 1.35)

= 0.4362 + 0.0019

= 0.4381 Answer

DONE