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pav-90 [236]
3 years ago
5

A random sample of n observations is selected from a normal population to test the null hypothesis that muμequals=10. Specify th

e rejection region for each of the following combinations of HSubscript aa​, alphaα​, and n. a. HSubscript aa​: muμnot equals≠​10; alphaαequals=0.010.01​; nequals=1313 b. HSubscript aa​: muμgreater than>​10; alphaαequals=0.100.10​; nequals=2323 c. HSubscript aa​: muμgreater than>​10; alphaαequals=0.050.05​; nequals=99 d. HSubscript aa​: muμless than<​10; alphaαequals=0.100.10​; nequals=1111 e. HSubscript aa​: muμnot equals≠​10; alphaα equals=0.050.05​; nequals=2020 f. HSubscript aa​: muμless than<​10; alphaαequals=0.010.01​; nequals=77 a. Select the correct choice below and fill in the answer box within your choice.
Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

Step-by-step explanation:

a) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 13

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a two tailed test. Therefore, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

The critical value is 3.012

The rejection region is area > 3.012

b) Ha: μ > 10

This is a right tailed test

n = 23

α = 0.1

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.319

The rejection region is area > 1.319

c) Ha: μ > 10

This is a right tailed test

n = 99

α = 0.05

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.05 = 0.95

The critical value is 1.66

The rejection region is area > 1.66

d) Ha: μ < 10

This is a left tailed test

n = 11

α = 0.1

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.363

The rejection region is area < 1.363

e) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 20

Since α = 0.05, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

The critical value is 2.086

The rejection region is area > 2.086

f) Ha: μ < 10

This is a left tailed test

n = 77

α = 0.01

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.01 = 0.99

The critical value is 2.376

The rejection region is area < 2.376

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Answer:

2x^2 + 7x + 3 is the area of the rectangular piece of wood.

Step-by-step explanation:

Here,

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3 years ago
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The identification of parts A,B andC is illustrated below with their various reasons given.

<h3>What is an equilateral triangle?</h3>

An equilateral triangle is the triangle that has all its sides equal in length and each angle is made up of angle 60°.

Part A = The similar triangle are RGE and PBE

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Part C= All the sides of the equilateral triangle are the same therefore the distance from B to E and from P to E is the same with BP which is 225ft.

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2 years ago
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4 0
3 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
12345 [234]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

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Luba_88 [7]

Answer:

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Step-by-step explanation:

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