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3 years ago

Explain how the digit 7 can have different values

Mathematics

1 answer:

kakasveta [241]3 years ago

7 0

It could be 7 groups of 1 3 groups of 2 with one left over or 2 groups of 3 with 1 left over

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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

What is the value of the x variable in the solution to the following system of equations?

harkovskaia [24]

Its A. x=4

procedure: 3*equation1 - 2*equation2 ⇒ x=4

8 0

3 years ago

Write 6 17/20 as a decimal

djverab [1.8K]

Answer:

6.85

Step-by-step explanation:

6 17/20

Convert to improper fraction.

137/20

Divide.

= 6.85

3 0

3 years ago

Read 2 more answers

Is 6 divisible by 36

andrew-mc [135]

In the terms you used, not unless you want an integer for an answer.

8 0

3 years ago

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PLS HELP. Find the first four terms of the recursive sequence defined by the following formula:

Gnoma [55]

Answer:

see explanation

Step-by-step explanation:

Given the recursive formula $a_{n}$ = $\frac{a_{n-1} }{4}$ and

$a_{4}$ = 2 $\frac{1}{4}$ = $\frac{9}{4}$ , then

$\frac{a_{3} }{4}$ = $\frac{9}{4}$ ( multiply both sides by 4 )

a₃ = 4 × $\frac{9}{4}$ = 9

$\frac{a_{2} }{4}$ = 9 ( multiply both sides by 4 )

a₂ = 36

$\frac{a_{1} }{4}$ = 36 ( multiply both sides by 4 )

a₁ = 144

The first 4 terms are

144, 36, 9, 2 $\frac{1}{4}$

3 0

3 years ago