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3 years ago

Given the expression 12C7, which of the following scenarios could this combination represent?

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer: B

<u>Step-by-step explanation:</u>

₁₂C₇ is read as "12 choose 7" which means you are choosing 7 out of 12. Since it is a combination and not a permutation, order does not matter.

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3 years ago

A recent study asked U. S. Adults to name 10 historic events that occurred during their lifetime that have had the greatest impa

Nadusha1986 [10]

Using the z-distribution, as we are working with a proportion, it is found that the 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, the parameters are:

$z = 2.575, n = 2000, \pi = 0.77$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942$

The 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

More can be learned about the z-distribution at brainly.com/question/25890103

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2 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.

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3 years ago