Classify scatter plot

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

Luke's Auto Wash earned $10 from servicing 2 customers. How many

laiz [17]

5 more because 10 divided by 2 is 5 and when u add 5 by 5 you get 10 and when you also add another 5 you get 15 so yeah

4 0

3 years ago

Please help!! Due really soon!

Alisiya [41]

9514 1404 393

Answer:

f(x) = 6x +1

Step-by-step explanation:

Differences in x-values (first row) are 1, 1, 1.

Differences in y-values (second row) are 6, 6, 6.

The constant ratio of differences (6/1) tells you the function is linear, and has a slope of m = 6/1 = 6.

Using the first point in the form ...

y = mx + b

we have ...

y = 6x + b

7 = 6·1 + b . . . . (x, y) = (1, 7)

1 = b . . . . . . subtract 6

Then the equation can be written ...

y = 6x +1

In functional form, this is ...

f(x) = 6x +1

4 0

3 years ago

What does "sum" mean in math?

Alexxx [7]

In math the sum means to add

6 0

3 years ago

Read 2 more answers

Is there enough information to prove P is parallel to Q if so state the theorem or postulate?

stira [4]

Yes there is. i do not know the theorem but I know it's rule. If 2 lines intersect the same line, and they have an equal angle in the same quadrant then they are parallel. This is because their all of their angles would be equal and they are basically the same line but shifted over

3 0

3 years ago

Read 2 more answers