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Gennadij [26K]
3 years ago
8

Setting:

Mathematics
2 answers:
wariber [46]3 years ago
7 0

a. No because he only surveyed those who were leaving and not a whole set.

b. It sounds biased because he uses the word "<em>lengthy</em>" which means long, and sounds opinionated.

c.  54-51 = 3

3 divided by 54 = 0.055

0.055 x 100 = 5.5

nata0808 [166]3 years ago
6 0
A. Mr. Kent interviewed the 54 students as they are going to leave the school, it is not considered to be a random sample. It is because a random sample is when a set is taken from a population. Mr. Kent interviewed the 54 who are going to leave, meaning, he didn't take a set out of that 54, he took all of them. So it is not a random sample.

b. The question that Mr. Kent asked is considered to be a leading question, so it does not seem biased.

c. If there are 54 respondents.
51 = yes, the rest is no.
= 54 - 51 = 3
= 3 is now divided to 54 = 3/54
= giving an answer of 0.0555
= 0.0555 x 100
= 5.6%
= The percent of responses that says 'no' is 5.6%

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A researcher used a sample of n = 60 individuals to determine whether there are any preferences among six brands of pizza. Each
Blizzard [7]

Answer:

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.

3) e_{females.} = 80

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

5) χ²_{6}

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}

I hope you have a SUPER day!

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