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3 years ago

13

13.225 + 2.6 what is the answer to that question

2 answers:

devlian [24]3 years ago

8 0

15.825 , your basically adding it up regularly instead you add a decimal !

grin007 [14]3 years ago

7 0

15.825 would be the answer to this

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-x/12 = -(1/12)(x) or D

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3 years ago

The vertices of a triangle are (-1.3) (-5,3) (-5,7) how would you reflect the triangle in the x-axis and in the y-axis ? Give th

grandymaker [24]

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3 years ago

A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that

mafiozo [28]

Answer:

(a) The significance level of the test is 0.002.

(b) The power of the test is 0.3487.

Step-by-step explanation:

We are given that a coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that the probability is not 0.5.

The test rejects the null hypothesis if either 0 or 10 heads are observed.

Let p = <u><em>probability of obtaining head.</em></u>

So, Null Hypothesis, $H_0$ : p = 0.5

Alternate Hypothesis, $H_A$ : p $\neq$ 0.5

(a) The significance level of the test which is represented by $\alpha$ is the probability of Type I error.

Type I error states the probability of rejecting the null hypothesis given the fact that the null hypothesis is true.

Here, the probability of rejecting the null hypothesis means we obtain the probability of observing either 0 or 10 heads, that is;

P(Type I error) = $\alpha$

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

Also, the event of obtaining heads when a coin is thrown 10 times can be considered as a binomial experiment.

So, X ~ Binom(n = 10, p = 0.5)

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

$\binom{10}{0}\times 0.5^{0} \times (1-0.5)^{10-0} +\binom{10}{10}\times 0.5^{10} \times (1-0.5)^{10-10}$ = $\alpha$

$(1\times 1\times 0.5^{10}) +(1 \times 0.5^{10} \times 0.5^{0})$ = $\alpha$

$\alpha$ = 0.0019

So, the significance level of the test is 0.002.

(b) It is stated that the probability of heads is 0.1, and we have to find the power of the test.

Here the Type II error is used which states the probability of accepting the null hypothesis given the fact that the null hypothesis is false.

Also, the power of the test is represented by (1 - $\beta$ ).

So, here, X ~ Binom(n = 10, p = 0.1)

$1-\beta$ = P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true)

$1-\beta$ = $\binom{10}{0}\times 0.1^{0} \times (1-0.1)^{10-0} +\binom{10}{10}\times 0.1^{10} \times (1-0.1)^{10-10}$

$1-\beta$ = $(1\times 1\times 0.9^{10}) +(1 \times 0.1^{10} \times 0.9^{0})$

$1-\beta$ = 0.3487

Hence, the power of the test is 0.3487.

3 0

3 years ago

Rewrite using powers of ten 32,000,000

mamaluj [8]

32×10×10×10×10×10×10

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3 years ago

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Evaluate 805 divide by 5

Ket [755]

805 divided by 5 equals 161

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3 years ago

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