A famous mathematician and physicist named Archimede was born in 287 B.C. Write an integer to express the year of his birth.

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

4 0

4 years ago

How is the square root calculated using the Newton-Raphson method?

DochEvi [55]

Step-by-step explanation:

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7 0

3 years ago

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Identify the function as a power function, a polynomial function, or neither.<br><br> f(x)=4(x^3)^3

Alborosie

Answer:7