What conclusion is appropriate if a chi-square test produces a chi-square statistic near zero?

1 answer:

5 0

<span>expected values are close to observed values, tend to favor the null hypothesis; existence of relationship is not supported, or like, departure from expectation is not supported.</span>

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A triangle has one angle measuring 3X degrees a second angle measures 2X +20°

Lena [83]

Answer : x = 20

Explanation :
3x + 2x +20 + 4x - 20 = 180

(3+2+4=9)

9x + 20 - 20
The 20s cancel each other out because one is positive and the other is negative

9x = 180

9x/9 = 180/9

X = 180 divided by 9

X = 20

7 0

3 years ago

The cost (in dollars) of a custom made sweatshirt is represented by 3.5n +29.99, where n is the number of different colors in th

juin [17]

Answer:

$52.5n+449.85$

Step-by-step explanation:

The cost is given for <em>1 sweatshirt</em>. To find the cost of 15 sweatshirts, we need to multiply the expression by 15. Let's do this:

$15(3.5n+29.99)\\=15(3.5n+29.99)\\=15(3.5n)+15(29.99)\\=52.5n+449.85$

This is the expression that represents the cost of 15 sweatshirts.

8 0

3 years ago

What's the answer to this? And how do u do it

andrew-mc [135]

-5n - 3 - 3 > 19
-5n - 6 > 19
+ 6 +6
-5n > 25
divide by -5 on each side
change signs since dividing by negative

N< -5

For graphing on the line, you will create an OPEN circle on -5 and sketch it to the left.

3 0

4 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$