What multiples to -36 but adds to 16
2 answers:
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Answer:
Andrew walks more steps than Sarah.
Step-by-step explanation:
The question is incomplete. The complete question is :
Andrew and Sarah are tracking the number of steps they walk. Andrew records that he can walk 6000 steps in 50 minutes. Sarah writes the equation y=118x, where y is the number of steps and x is the number of minutes she walks, to describe her step rate. This week, Andrew and Sarah each walk for a total of 5 hours. Who walks more steps?
Solution:
Given:
Andrew walks 6000 steps in 50 minutes.
Number of steps Sarah walks is given by the equation :
where is the number of steps and
is the number of minutes she walks.
Finding the number of steps each walks in 5 hours.
Total number of minutes in 5 hours =
For Andrew:
Using unitary method:
If in 50 minutes Andrew walks = 6000 steps.
In 1 minute he will walk = = 120 steps
In 300 minutes he will walk = steps
Thus, Andrew will walk 36,000 steps in 5 hours.
For Sarah:
We will plugin in the equation and solve for
.
We have:
∴
Thus, Sarah will walk 35,400 steps in 5 hours.
Therefore, Andrew walks more steps than Sarah as .
Answer:
1) χ² ≥ 11.07
2) Goodness of fit test, df: χ²
Independence test, df: χ²
The goodness of fit test has more degrees of freedom than the independence test.
3)
4) H₀: ∀ i= 1, 2, ..., r and j= 1, 2, ..., c
5) χ²
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= ∑ ≈χ²
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
χ²
χ² = 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= ∑ ≈χ²
Degrees of freedom: χ²
In the example: k= 4 (the variable has 4 categories)
χ² = χ²
The statistic for the independence test is:
χ²= ∑∑ ≈χ²
∀ 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
χ² = χ²
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.
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)
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: ∀ 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= ∑∑ ≈χ²
∀ i= 1, 2, ..., r & j= 1, 2, ..., c
In the example: c= 3 and r= 4
Degrees of freedom: χ²
χ² = χ²
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