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katen-ka-za [31]
3 years ago
8

If the critical z-value for a hypothesis test... If the critical z-value for a hypothesis test equals 2.45, what value of the te

st statistic would provide the least chance of making a Type I error?
Mathematics
1 answer:
schepotkina [342]3 years ago
8 0

Answer: z score = 0.00714

Step-by-step explanation: the value of test statistics is gotten using the standard normal distribution table.

Z= 2.45 has area to the left (z<2.45) and area to the right (z>2.45).

Level of significance α is the probability of committing a type 1 error. The area under the distribution is known as the rejection region and it is the area towards the right of the distribution.

The table I'm using is towards the left of the distribution.

But z>2.45 + z<2.45 = 1

z> 2.45 = 1 - z<2.45

But z < 2.45 = 0.99286

z > 2.45 = 1 - 0.99286

z >2.45 = 0.00714

Hence the test statistics that would produce the least type 1 error is 0.00714

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x²-12x+36

Step-by-step explanation:

an expression in the form (a-b)² is expanded to the form a²-2ab+b²

you can also expand it to (x-6)×(x-6) then multiply the first term by the first the outer term by the outer term the inner term by the inner term and the last term by the last term (foil)

the first terms are x and x = x² the outer terms are x and -6 = -6x the inner terms are also x and -6 = -6x

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adding all the four products =

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3/4 + 1/2
multiply 1/2 denominator and numerator by 2 to match 3/4
= 3/4 + 2/4 = 5/4 (copy same denominator add numerator)

2/6 + 1/3
divide 2/6 denominator and numerator by 2 to match 1/3
= 1/3 + 1/3 = 2/3 (copy same denominator add numerator)

5/9 + 2/3
multiply 2/3 denominator and numerator by 3 to match 5/9
= 5/9 + 6/9 = 11/9 (copy same denominator add numerator)



6/9 -1/5
cross multiply 6x5 - 1x9 for numerator
for denominator multiply 9x5
=30/45 - 9/45= 21/45
divide num and den by 3
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5/8-1/3
cross multiply 5x3-1x8 for numerator
multiply 8x3 for denominator
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