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alexandr402 [8]
3 years ago
15

Consumers Energy states that the average electric bill across the state is $67.16. You want to test the claim that the average b

ill amount is greater than $67.16. The hypotheses for this situation are Null Hypothesis: μ ≤ 67.16, Alternative Hypothesis: μ > 67.16. If the true statewide average bill is $51.28 and the null hypothesis is not rejected, did a type I, type II, or no error occur?
a. Type I Error has occured.
b. No error has occured.
c. We do not know the degrees of freedom, so we cannot determine if an error has occured.
d. Type II Error has occured
e. We do not know the p-value, so we cannot determine if an error has occured.
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

B: No error has occured

Step-by-step explanation:

In statistical hypothesis testing, we have two types of errors namely Type I error and Type II error.

Type I error occurs when we reject the null hypothesis although it is true. Meanwhile, Type II error occurs when we fail to reject the null hypothesis null hypothesis even though it is false.

Now, we are given the null and alternative hypothesis as;

Null Hypothesis: μ ≤ 67.16

Alternative Hypothesis: μ > 67.16.

Where μ represents the statewide average bill.

We are also given the true statewide average bill is $51.28.

Now, the true statewide average bill falls within the range of the null hypothesis given.

Thus, we can't reject the null hypothesis since it is true.

Now, we are told the null hypothesis is not rejected and that is correct.

Thus, there is no error as it doesn't fall into the type I or the type II error.

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3 years ago
Determine whether each of these functions from Z to Z is one-to-one. a) f(n) = n - 1 b) f(n) = n2 + 1 c) f(n) = nº d) f(n) = [n/
sesenic [268]

Answer:  The correct option is

(a) f(n) = n - 1.

Step-by-step explanation:  We are given to determine whether the given functions are one-to-one or not.

We know that a function y = f(x) is one-to-one if and only if

f(x) = f(y)  ⇔  x = y.

That is, any two distinct elements cannot have the same image.

(a) The given function is

f(n)=n-1.

Let us consider that

f(n_1)=f(n_2)\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow n_1=n_2.

Similarly,

n_1=n_2\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow f(n_1)=f(n_2).

So, this function is one-to-one.

(b) The given function is

f(n)=n^2+1.

Let us consider that

f(n_1)=f(n_2)\\\\\Rightarrow n_1^2+1=n_2^2+1\\\\\Rightarrow n_1^2=n_2^2\\\\\Rightarrow n_1=\pm n_2.

That is, there may be two unequal elements having same image.

For example, f(-1)=(-1)²+1=1+1=2,  f(1)=(1)²+1=1+1=2.

It implies that f(-1)=f(1) but 1 ≠ -1.

So, the given function is not one-to-one.

(c) The given function is

f(n)=n^0.

Here, the image of all the elements is 1.

For example, f(2)=2^0=1,~~f(3)=3^0=1.

f(2)=f(3)  but  2≠3.

So, more than one element is having the same image and so the function cannot be one-to-one.

(d) The given function is

f(n)=\left[\dfrac{n}{2}\right].

Here, we see that

f(2)=\left[\dfrac{2}{2}\right]=[1]=1,\\\\\\f(3)=\left[\dfrac{3}{2}\right]=[1.5]=1.

So, f(2)=f(3) but 2≠3.

So, the given function is not one-to-one.

Thus, the correct option is (a).

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Situation of non- proportional relationship :

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