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Basile [38]
3 years ago
8

When we test Upper H 0​: muequals0 against Upper H Subscript a​: mugreater than​0, we get a​ P-value of 0.03. a. What would the

decision be for a significance level of 0.10​? Interpret in context. b. If the decision in​ (a) is in​ error, what type of error is​ it? c. Suppose the significance level were instead 0.01. What decision would you​ make, and if it is in​ error, what type of error is​ it?
Mathematics
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

(a) The null hypothesis will be rejected.

(b) Type I error

(c) The null hypothesis will not be rejected. The error is type II error.

Step-by-step explanation:

The hypothesis provided is:

H_{0}: \mu = 0\\ H_{a} : \mu > 0

The p-value of the test obtained is 0.03

(a)

Decision rule for hypothesis testing, based on p-value, states that if the p-value is less than the significance level (α) then the null hypothesis is rejected and vice versa.

The significance level is α = 0.10

Then,

p-value = 0.03 < \alpha  = 0.10

Thus, the null hypothesis will be rejected.

Conclusion:

The null hypothesis is rejected stating that the value of μ is more than 0.

(b)

If the decision in​ (a) is an​ error, i.e. the null hypothesis is rejected when in fact it is true, this type of error is known as type I error.

(c)

The significance level is α = 0.01

Then,

p-value = 0.03 > \alpha  = 0.01

Thus, the null hypothesis will not be rejected.

Conclusion:

The null hypothesis was not rejected stating that the value of μ is 0.

If this decision is an error, i.e. the null hypothesis was not rejected when in fact it is false, this type of error is known as type II error.

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What is the area of the figure? A) 18 in2 B) 24 in2 C) 30 in2 D) 36 in2
Annette [7]
Hello!

You need to separate this into two rectangles and add their areas together

first rectangle

3 * 6 = 18

rectangle 2

3 * 2 = 6

18 + 6 = 24

the answer is 24in squared


5 0
3 years ago
Read 2 more answers
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
3 years ago
How are angles named?
Mama L [17]

Answer:

Acute angle, right angle, obtuse angle and reflex angle.

Step-by-step explanation:

Acute angle -

0° < θ < 90°

Right angle -

θ = 90°

Obtuse angle -

90° < θ < 180°

Reflex angle -

θ > 180°

4 0
3 years ago
A square with sides of 3 squareroot 2 is inscribed in a circle. what is the area of one of the sectors formed by the radii to th
mrs_skeptik [129]
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2).  Draw this to verify this statement.  Note that the height of each such triangular area is (3 sqrt(2))/2.

So now we have the base and height of one of the triangular sections.

The area of a triangle is A = (1/2) (base) (height).  Subst. the values discussed above,   A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2).  Show that this boils down to A = 9/2.

You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section.   Doing the problem this way, we get (1/4) (3 sqrt(2) )^2.  Thus, 
A = (1/4) (9 * 2) = (9/2).  Same answer as before. 
4 0
3 years ago
Luke and Matthew ran a lemonade stand on Saturday. They agreed that Matthew would get 60%, percent of the profit because the lem
Svetlanka [38]
Matthew made $15


Matthew 60% + Luke 40%= Profit 100%


if profit is $25; 60% will be: 0.60* $25=$15 ; Matthew's share 
                       40% will be:  0.40* $25 =10 ; Luke share         
3 0
3 years ago
Read 2 more answers
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