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exis [7]
3 years ago
7

If a hypothesis is not rejected at the 0.10 level of significance, it: a. may be rejected at the 0.05 level. b. must be rejected

at the 0.025 level. c. must be rejected at the 0.05 level. d. will not be rejected at the 0.05 level.
Mathematics
1 answer:
mariarad [96]3 years ago
4 0

Answer:

Option D - Will not be rejected at the 0.05 level.

Step-by-step explanation:

The significance level, which is denoted as "α", is a measure of the strength of the evidence that must be present in a sample before we can reject the null hypothesis and conclude that the effect is statistically significant. Now, this significance level must be determined before conducting an experiment.

Now, in the context of this question, the significance level is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 means a 5% risk of concluding that a difference exists when there is no actual difference. Now, lower significance levels will indicate that we require stronger evidence before we can reject the null hypothesis.

Thus, if we don't reject at α = 0.1,we obviously will not reject at higher values.

Thus, looking at the options, we will not reject at 0.05 significance level.

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3 years ago
Simplify the expression: 4(x + 3) + 3(5 – x) A X + 27 B X + 8 C 2x + 15 D. 19x + 11​
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Read 2 more answers
Consider the following function. f(x) = 7x + 1 x (a) Find the critical numbers of f. (Enter your answers as a comma-separated li
sukhopar [10]

Answer:

a) DNE

b) The function increases for every real value of x.

c) DNE

Step-by-step explanation:

Given a function f(x), the critical points are those in which f^{\prime}(x) = 0, that is, the roots of the first derivative of f(x).

Those critical points let us find the intervals in which the function increases or decreases. If the first derivative in the interval is positive, the function increases in the interval. If it is negative, the function decreases.

If the function increases before a critical point and then, as it passes the critical point, it starts to decrease, we have that the critical point (x_{c}, f(x_{c}) is a relative maximum.If the function decreases before a critical point and then, as it passes the critical point, it starts to increase, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative minimum.If the function has no critical points, it either always increases or always decreases.In this exercise, we have that:[tex]f(x) = 7x + 1

(a) Find the critical numbers of f.

f^{\prime}(x) = 0

f^{/prime}(x) = 7

7 = 0 is false. This means that f has no critical points.

(b) Find the open intervals on which the function is increasing or decreasing.

Since there are no critical points, we know that either the function increases or decreases in the entire real interval.

We have a first order function in the following format:

f(x) = ax + b

In which a > 0.

So the function increases for every real value of x.

(c) Apply the First Derivative Test to identify the relative extremum.

From a), we find that there are no critical numbers. So DNE

7 0
3 years ago
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