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Eva8 [605]
3 years ago
9

The isotope samarium-151 decays into europium-151, with a half-life of around 96.6 years. A rock contains 5 grams of samarium-15

1 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.
The time since the rock reached its closure temperature is _____


years. When the rock was discovered, it had _____


grams of europium-151.
Mathematics
1 answer:
Olegator [25]3 years ago
7 0
Part A:

The amount of substance left of a radioactive substance with a half life t_{ \frac{1}{2} } and amount of initial substance N_0 is given by

N(t)=N_0\left( \frac{1}{2} \right)^{ \frac{t}{t_{ \frac{1}{2} }} }

Given that th<span>e isotope samarium-151 has a half-life of around 96.6 years and the rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.

</span><span>The time since the rock reached its closure temperature is obtained as follows:

0.625=5\left( \frac{1}{2} \right)^{ \frac{t}{96.6} } \\  \\ \Rightarrow\left( \frac{1}{2} \right)^{ \frac{t}{96.6} }= \frac{0.625}{5} =0.125 \\  \\ \Rightarrow\frac{t}{96.6}\log{ \frac{1}{2} }=\log0.125 \\  \\ \Rightarrow\frac{t}{96.6}= \frac{\log0.125}{\log{ \frac{1}{2} }} =3 \\  \\ \Rightarrow t=3(96.6)=289.8

Therefore, </span>t<span>he time since the rock reached its closure temperature is 289.8 years.



Part B:

Given that the </span><span>rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered. The number of grams of europium-151 when the rock was discovered is 5 - 0.625 = 4.375</span>
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1) χ² ≥ 11.07

2) Goodness of fit test, df: χ²_{3}

Independence test, df: χ²_{1}

The goodness of fit test has more degrees of freedom than the independence test.

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Step-by-step explanation:

Hello!

1)

The researcher took a sample of n=60 people and made them taste proof samples of six different brands of pizza and choose their favorite brand, their choose was recorded. So the study variable is the following:

X: favorite pizza brand, categorized in brand 1, brand 2, brand 3, brand 4, brand 5 and brand 6.

The Chi-square goodness of fit test is done with the following statistic:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Where k represents the number of categories of the study variable. In this example k= 6.

Remember, the rejection region for the Chi-square tests of "goodnedd of fit", "independence", and "homogeneity" is allways one-tailed to the right. So you will only have one critical value.

χ²_{k-1; 1 - \alpha }

χ²_{6-1; 1 - 0.05 }

χ²_{5; 0.95 } = 11.070

This means thar the rejection region is χ² ≥ 11.07

If the Chi-Square statistic is equal or greather than 11.07, then you reject the null hypothesis.

2)

The statistic for the goodness of fit is:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Degrees of freedom: χ²_{k-1}

In the example: k= 4 (the variable has 4 categories)

χ²_{4-1} = χ²_{3}

The statistic for the independence test is:

χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

If the information is in a contingency table

r= represents the total of rows

c= represents the total of columns

In the example: c= 2 and r= 2

Degrees of freedom: χ²_{(r-1)(c-1)}

χ²_{(2-1)(2-1)} = χ²_{1}

The goodness of fit test has more degrees of freedom than the independence test.

3)

To calculate the expected frecuencies for the independence test you have to use the following formula.

e_{ij} = n * P_i. * P_.j = n * \frac{o_i.}{n} * \frac{o_.j}{n}

Where o_i. represents the total observations of the i-row, o_.j represents the total of observations of the j-column and n is the sample size.

Now, this is for the expected frequencies in the body of the contingency table, this means the observed and expected frequencies for each crossing of categories is not the same.

On the other hand, you would have the totals of each category and population in the margins of the table (subtotals), this is the same when looking at the observed frequencies and the expected frequencies. Wich means that the expected frequency for the total of a population is the same as the observed frequency of said population. A quick method to check if your calculations of the expected frequencies for one category/population are correct is to add them, if the sum results in the subtotal of that category/population, it means that you have calculated the expected frequencies correctly.

The expected frequency for the total of females is 80

Using the formula:

(If the females are in a row) e_{females.} = 100 * \frac{80}{100} * \frac{0}{100}

e_{females.} = 80

4)

There are two ways of writing down a null hypothesis for the independence test:

Way 1: using colloquial language

H₀: The variables X and Y are independent

Way 2: Symbolically

H₀: P_{ij}= P_{i.} * P_{.j} ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

This type of hypothesis follows from the definition of independent events, where if we have events A and B independent of each other, the probability of A and B is equal to the product of the probability of A and the probability of B, symbolically: P(A∩B) = P(A) * P(B)

5)

In this example, you have an independence test for two variables.

Variable 1, has 3 categories

Variable 2, has 4 categories

To follow the notation, let's say that variable 1 is in the rows and variable 2 is in the columns of the contingency table.

The statistic for this test is:

χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

In the example: c= 3 and r= 4

Degrees of freedom: χ²_{(r-1)(c-1)}

χ²_{(3-1)(4-1)} = χ²_{6}

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