An article in Food Testing and Analysis measured the sugar concentration in (10)

Question

An article in Food Testing and Analysis measured the sugar concentration in (10) clear apple juice. All readings were taken at the same temperature. Assume that the sugar concentration in this type of clear apple juice is normally distributed with known standard deviation equal to 0.2. Suppose that we want to test Ho = 11.5 against Hi: 1300. Let X denote the sample average compressive strength for n 20 randomly selected specimens. Let >1331.26 be the rejection region. Suppose compressive strength for specimens is normally distributed with = 60, what is the probability of type I error.

Explanation / Answer

2.a.
i.
Given that,
Standard deviation, =0.2
Sample Mean, X =11.4
Null, H0: =11.5
Alternate, H1: <11.5
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-11.5)/0.2/(n) < -1.6449 OR if (x-11.5)/0.2/(n) > 1.6449
Reject Ho if x < 11.5-0.329/(n) OR if x > 11.5-0.329/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 36 then the critical region
becomes,
Reject Ho if x < 11.5-0.329/(36) OR if x > 11.5+0.329/(36)
Reject Ho if x < 11.445 OR if x > 11.555
Implies, don't reject Ho if 11.445 x 11.555
Suppose the true mean is 11.4
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(11.445 x 11.555 | 1 = 11.4)
= P(11.445-11.4/0.2/(36) x - / /n 11.555-11.4/0.2/(36)
= P(1.35 Z 4.65 )
= P( Z 4.65) - P( Z 1.35)
= 1 - 0.9115 [ Using Z Table ]
= 0.089
For n =36 the probability of Type II error is 0.089

ii.
power of the test = 0.9
power of the test = 1-beta =0.9
beta =(type2 error) = 0.1
sample size = n=?
n= (((Zalpha +Zbeta))/(U-Uo))^2
Z alpha at 0.05 = 1.96
Z beta at 0.10 = 1.28
n= ((0.2(1.96+1.28))/(11.5-11.4)^2
n =64.8= 65

2.b.
Given that,
Standard deviation, =60
Sample Mean, X =1331.26
Null, H0: =1300
Alternate, H1: >1300
Level of significance, = 0.05
From Standard normal table, Z /2 =1.645
Since our test is right-tailed
Reject Ho, if Zo < -1.645 OR if Zo > 1.645
Reject Ho if (x-1300)/60/(n) < -1.645 OR if (x-1300)/60/(n) > 1.645
Reject Ho if x < 1300-98.7/(n) OR if x > 1300-98.7/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 1300-98.7/(20) OR if x > 1300+98.7/(20)
Reject Ho if x < 1277.93 OR if x > 1322.07
Suppose the true mean is 1331.26
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(1277.93 < x OR x >1322.07 | 1 = 1331.26)
= P(1277.93-1331.26/60/(20) < x - / /n OR x - / /n >1322.07-1331.26/60/(20)
= P(-3.975 < Z OR Z >-0.685 )
= P( Z <-3.975) + P( Z > -0.685)
= 0 + 0.7533 [ Using Z Table ]
= 0.753

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