Mathematics, 30.07.2019 23:30 Geo777

You suspect that an unscrupullous employee at a casino has tampereed with a die. that is, he is using a loaded die. in order to test this claim you roll it 258 times and obtain the folllowing frquencies category 1 2 3 4 5 6 frequency 48 52 51 34 48 25 a. choose the approp. alternative hypotheses to test if the population proportion differ a/all population proportion differ from 1/6 b/ not ll population proportion are equal to 1/6 b-1 calculate the value of the test statistic(round to at last 4 decimal places) b-2 approx. the p-value c. at the 10% signifigance level. can you conclude that the die is loaded?

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Answer from: bburpee1

A. For the alternative hypothesis, you normally would want to choose something that says not equal since you will need to be testing if the proportions are equal and this will be your null hypothesis.

Therefore for your alternative hypothesis it would need to be: Not all population proportions are equal to 1/6.

b-1. Since we have 6 different outcomes, we will use the chi square test to evaluate the hypothesis. The test statistic is computed by getting the summation of $\frac{ (observed-expected)^{2} }{expected}$ for all the 6 outcomes. In our case the expected proportion would be 1/6 or 0.1667 and the observed values would be the proportions from the 6 outcomes.

$X^{2} = \frac{ (0.1860-0.1667)^{2} }{0.1667} +\frac{ (0.2016-0.1667)^{2} }{0.1667}+\frac{ (0.1977-0.1667)^{2} }{0.1667}$
$+\frac{ (0.1318-0.1667)^{2} }{0.1667}+\frac{ (0.1860-0.1667)^{2} }{0.1667}+\frac{ (0.0969-0.1667)^{2} }{0.1667}=0.0541$

ANSWER: The test statistic is 0.0541

b-2. For this problem we just imagine the chi square distribution and the fact that our test statistic is so close to zero so the p-value must be so much greater than 0.10.

Based on other resources, there are choices to this problem so we just choose the option similar to the one we just described.

ANSWER: p-value > 0.10

c. Since the p-value we estimated above is greater than 0.10, we do not reject the null hypothesis. This means that there is NOT enough evidence to reject the fact that the die is a fair die.

ANSWER: No, we cannot conclude that the die is loaded.

Answer from: Quest

Where’s the graph too show me?

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

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You suspect that an unscrupullous employee at a casino has tampereed with a die. that is, he is usin...