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2 years ago

No chi-square The principal in Exercise 7 also

Mathematics

1 answer:

larisa [96]2 years ago

8 0

The random sample of the students is an illustration of sampling

The chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.

<h3>How to determine the reason chi square is not appropriate?</h3>

The dataset is given as:

Monday 34

Tuesday 29

Wednesday 32

Thursday 28

Friday 19

The variable of the above dataset is a not a categorical dataset.

One of the conditions of the chi-square test for goodness of fit test is that the variable under study must be categorical.

Hence, the chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.

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the numbers of girls in the choir is 4 times the number of boys in the choir .the total number of students is 60..how many girls

Inessa05 [86]

First, make up some variables to represent the number of Girls and Boys in the choir.

B = number of boys
G = number of girls

You know that there are 4 times as many girls in the choir as boys. Therefore, the equation you can write is:
$\frac{B}{G} = \frac{1}{4}$

If you cross-multiply, then you get the simplified equation:
G = 4B

Intuitively this makes sense since if you multiplied the number of boys in the class by 4, that would be equal to the number of girls you have.

Now, we know that the total class size is 60. So girls plus boys equals 60:
G+B = 60

To solve the equation, replace the G in this equation with the replacement you found before, 4B.
G + B = 60 -->
4B + B = 60 -->
5B = 60 -->
B = 12

However, you are trying to find the number of girls, so plug the answer back into your equation.
G + B = 60 -->
G + 12 = 60 -->
G + 12 -12 = 60 - 12 -->
G = 48

The number of girls you have is 48.

6 0

3 years ago

Which sign makes the statement true? 93/100 ? 2/5

polet [3.4K]

Answer:

Step-by-step explanation:

93/100 = 0.93

2/5 = 0.4, so:

93/100 > 2/5.

7 0

3 years ago

Explain what happens when you round 4.999 to the nearest tenth

Reptile [31]

It would round to 5 because 4.999 is closer to 5 than a decimal (idk if that's 100%)

5 0

3 years ago

The graph of a quadratic function does not cross the x-axis, but crosses the y-axis in one place.

natulia [17]

Answer: D

Step-by-step explanation:

A quadratic function has a degree of 2 so there will be two roots. The statement says the function does NOT CROSS THE X-AXIS so there are no real roots. That means both roots must be imaginary (complex).

5 0

3 years ago

Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair

Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102..

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

3*5: numbers with common factors >1, with 3*5 must be

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....} containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0

3 years ago