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sashaice [31]
3 years ago
12

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Mathematics
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

A.the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

\mathtt{H_o: \mu = 100}

\mathtt{H_1: \mu \neq 100}

A. If the acceptance region is defined as 98.5 <  \overline x >  101.5 , find the type I error probability \alpha .

Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

∴

\mathtt{\alpha = P( type  \ 1  \ error ) = P( reject \  H_o)}

\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5  )}

when  \mu = 100

\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

∴

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 103

\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 105

\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

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I'm not sure what this means. If you have choices you should list them.

(1/2)*(1/4 + 1/6) is an example of what should be given. There are two ways to solve this.

1. Use the distributive property.

1/2*1/4 + 1/2* 1/6

1/8 + 1/12 Which can be added using the LCD of 24

3/24 + 2/24 = 5/24

Method 2

Add what is inside the brackets first.

1/2 ( 1/4 + 1/6)

1/2(3/12 + 2/12 = 5/12

Now multiply by 1/2

1/2(5/12) = 5*1/(12 * 2) = 5 / 24 Same answer.

4 0
3 years ago
You accept a new job with a starting salary of $47,000. You receive a 4% raise at the start of your second year, a 5.6% raise at
Sunny_sXe [5.5K]

Answer:

1yr  47,000, 2nd yr  48,880, 3rd yr   51,617.28, 4th yr  57,346.7981 or 57,346.80

Step-by-step explanation:

1yr        47,000

2nd yr  47,000 x 4% = 48,880

3rd yr   48,880 x 5.6% = 51,617.28

4th yr   51,617.28 x 11.1% = 57,346.7981 or 57,346.80

4 0
3 years ago
Is the square root of -11.2 a rational number?
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No, it is an imaginary number.
Can't take the square root of a negative number without creating a method termed an imaginary number, which is the square root of -1 which they have used " i " to signify.
Then square root of -11.2 would be written Sqrt(11.2)i.  The Sqrt(11.2) = 3.3466
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3 years ago
In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro
VladimirAG [237]

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

<u>Step1:-</u>

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }

Given n=81 , σ =1.2 and  Zₐ = 1.96

<u>Step2:-</u>

<u />Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }<u />

<u />Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }<u />

On calculating , we get

Margin of error = 0.261

<u>Conclusion:-</u>

Margin of error  for a 95% of confidence intervals is 0.261

<u />

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3 years ago
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
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