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2 years ago

Two number cubes are rolled to determine how a token moves on a game board. The sides of

Mathematics

1 answer:

ahrayia [7]2 years ago

8 0

Answer:

D. The mathematical expectation of Option A is 1. The mathematical expectation of Option B is 1.5. Option B offers a greater likelihood of advancing to the finish line.

Step-by-step explanation:

The result of a product is odd only when the two numbers are odds.

There are 6*6 = 36 possible outcomes when two dice are rolled. Only 9 of them are a combination of two odd numbers: {1, 1} {1, 3} {1, 5} {3, 1} {3, 3} {3, 5} {5, 1} {5, 3} {5, 5}. Then 36 - 9 = 27 outcomes are even.

P(even) = 27/36 = 0.75

Option A) Mathematical expectation: 0.75*4 + 0.25*(-8) = 1

Option B) Mathematical expectation: 0.75*5 + 0.25*(-9) = 1.5

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A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.