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3 years ago

What is the difference between type i error and type ii error?

1 answer:

5 0

A type i error, as an example, is saying to a person that doesn't have ADHD, you have ADHD, or a false positive. A type ii error is saying to a person who does have ADHD, you don't have ADHD, or a false negative.

I hope this helped! :)

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Leona has 4 nickels, 6 pennies, 7 dimes and 8 quarters in a piggy bank. Leona’s sister Daisy picks a coin at random, record what

Radda [10]

Answer:

nvm.

Step-by-step explanation:

4 0

3 years ago

Use synthetic division to divide the polynomials. (2y^3 + 6y² + 7y - 6) / (y + 1)

uysha [10]

Answer:

Step-by-step explanation:

-1_| 2 6 7 -6

-2 -4 -3

2 4 3 -9

2y^2 + 4y + 3 - 9/(y + 1)

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3 years ago

1 of 12 Solve for X x + 15 = 21 X = = 0

natka813 [3]

Answer:

X=6

Step-by-step explanation:

To find the answer you could do 21 - 15 and that gets you an answer of 6 and thats how to get your answer.

7 0

3 years ago

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in in

Grace [21]

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

P.Q. = $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }$ = 5.063

$\sigma$ = population standard deviation

n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, $\sigma$ is ;

P(11.59 < $\chi^{2}__2_1$ < 32.67) = 0.90 {As the critical value of chi at 21 degrees

of freedom are 11.59 & 32.67}

P(11.59 < $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ < 32.67) = 0.90

P( $\frac{ 11.59}{(n-1) \times s^{2}}$ < $\frac{1}{\sigma^{2} }$ < $\frac{ 32.67}{(n-1) \times s^{2}}$ ) = 0.90

P( $\frac{(n-1) \times s^{2} }{32.67 }$ < $\sigma^{2}$ < $\frac{(n-1) \times s^{2} }{11.59 }$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1) \times s^{2} }{32.67 }$ , $\frac{(n-1) \times s^{2} }{11.59 }$ ]

= [ $\frac{21 \times 5.063^{2} }{32.67 }$ , $\frac{21 \times 5.063^{2} }{11.59 }$ ]

= [16.48 , 46.45]

90% confidence interval for $\sigma$ = [ $\sqrt{16.48}$ , $\sqrt{46.45}$ ]

= [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

5 0

3 years ago

Numbers 2-4 Will choose brainliest

Sever21 [200]

Answer:

2. 9, 8+9

3. 17-8=9, 17, 8, 9

4. 13-7=6

Step-by-step explanation:

6 0

3 years ago