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Nadya [2.5K]
3 years ago
7

If a researcher wants to examine the relationship among variables and not the difference, what test statistics should he or she

use?
Mathematics
1 answer:
BARSIC [14]3 years ago
5 0
There are various statistical tests which determine the differences between groups or more like anova, t-test, f-test, etc. examples on the other hand of test statistics that examine the relationship of variables includes linear regression which tests the r^2 and correlation.
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Supposethat z is a normally distributed variable with variance 4. You collect a sample of size n for which you get a sample aver
Nikitich [7]

Answer:

The sample size is 4

Step-by-step explanation:

Null hypothesis: The mean is 2

Alternate hypothesis: The mean is less than 2

Mean = 3

sd = sqrt(variance) = sqrt(4) = 2

At 95% confidence level, t-value is 1.960

Assuming the lower bound of the mean is 1.04

Lower bound = mean - (t×sd/√n)

1.04 = 3 - (1.96×2/√n)

3.92/√n = 3 - 1.04

3.92/√n = 1.96

√n = 3.92/1.96

√n = 2

n = 2^2

n = 4

7 0
3 years ago
Which term describes point U?
svlad2 [7]
The correct answer is B 

7 0
3 years ago
Portfolio Option 1: Sports Graph
Furkat [3]

Answer:The graph represent skydiving, Because when you jump out of the plane you have a bunch of speed and when you pull your parachute the speed will drop exceedingly.

Step-by-step explanation:

5 0
2 years ago
An automobile manufacturer has discovered that 20% of all the transmissions it installed in a particular style of truck are defe
Hatshy [7]

Answer:

0.148 = 14.8% probability that they will need to order at least one more new transmission

Step-by-step explanation:

For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

20% of all the transmissions it installed in a particular style of truck are defective after a year of use.

This means that p = 0.2

Sold seven trucks:

This means that n = 7

It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?

This is the probability that at least 3 are defective, that is:

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097

P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670

P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148

0.148 = 14.8% probability that they will need to order at least one more new transmission

6 0
3 years ago
Mark buys 3 2/5 pounds of apples. He uses 2/3 of the apples in a recipe. Which of the following expressions can be used to find
Dahasolnce [82]

Answer:

b

Step-by-step explanation:

jayfeather friend me

7 0
3 years ago
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