L(2) Time left 2:15:11 Let X1, X2,..., X, be a random sample of bacterial counts in n one ml random samples of your tap water, and suppose that the bacterial count in a ml of your tap water has a distribution with probability mass function: and 0 otherwise. e-80x ƒ (x|0) = I! if > 0 and x = 0, 1, 2, . . . . . It is desired to test Ho: 0=1 against H₁: 0= 2 at the 0.05 level of significance. The decision rule of the best test of the hypotheses is Select one: O A. -P(x-1) i-1 X; > c|0 = reject Ho if-n>e where c satisfies 0.05 = P OB. O C O'D. i1 12 reject Ho if εθ i-1 reject Ho if cwhere c satisfies 0.05 = PX; > c|0 i=1 i=1
L(2) Time left 2:15:11 Let X1, X2,..., X, be a random sample of bacterial counts in n one ml random samples of your tap water, and suppose that the bacterial count in a ml of your tap water has a distribution with probability mass function: and 0 otherwise. e-80x ƒ (x|0) = I! if > 0 and x = 0, 1, 2, . . . . . It is desired to test Ho: 0=1 against H₁: 0= 2 at the 0.05 level of significance. The decision rule of the best test of the hypotheses is Select one: O A. -P(x-1) i-1 X; > c|0 = reject Ho if-n>e where c satisfies 0.05 = P OB. O C O'D. i1 12 reject Ho if εθ i-1 reject Ho if cwhere c satisfies 0.05 = PX; > c|0 i=1 i=1
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:L(2)
Time left 2:15:11
Let X1, X2,..., X, be a random sample of bacterial counts in n one ml random samples of your tap water, and suppose
that the bacterial count in a ml of your tap water has a distribution with probability mass function:
and 0 otherwise.
e-80x
ƒ (x|0) =
I!
if > 0 and x = 0, 1, 2, . . . .
.
It is desired to test Ho: 0=1 against H₁: 0= 2 at the 0.05 level of significance.
The decision rule of the best test of the hypotheses is
Select one:
O A.
-P(x-1)
i-1
X; > c|0 =
reject Ho if-n>e where c satisfies 0.05 = P
OB.
O C
O'D.
i1
12
reject Ho if <c where c satisfies 0.05 =PX<c0
1-1
reject H₁ if - x; ≥ c where c satisfies 0.05 = P
TL
1-1
i=1
TE
TL
Σ Χ > εθ
i-1
reject Ho if cwhere c satisfies 0.05 = PX; > c|0
i=1
i=1
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