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rjkz [21]
3 years ago
15

A t statistic was used to conduct a test of the null hypothesis H0: µ = 11 against the alternative Ha: µ ? 11, with a p-value eq

ual to 0.042. A two-sided confidence interval for µ is to be considered. Of the following, which is the largest level of confidence for which the confidence interval will NOT contain 11?
a. A 90% confidence level
b. A 92% confidence level

c. A 96% confidence level
d. A 97% confidence level
e. A 98% confidence level
Mathematics
1 answer:
Andrews [41]3 years ago
5 0

Answer:

e. A 98% confidence level.

Step-by-step explanation:

Confidence level represents the percentage of probability as well as certainty that the confidence interval would contain the true population parameter when a random sample is selected many times.

Confidence intervals provide us with an upper and lower limit surrounding the sample mean, and within this interval we can then be confident we have captured the population mean.

From the information given above, to solve this question, we need to find the largest level of confidence for which the confidence interval will NOT contain 11.

In order to do this, we must produce the confidence interval.

Producing the confidence interval, for H0: µ = 11 against Ha: µ ?

11 would definitely produce a P-value less than (<) 0.042. The reason being that we can reject null hypothesis and confidence interval will not contain the value 11.

From given options above we have values as 0.10, 0.08, 0.04, 0.03, and 0.02.

Out of all these values, p value = 0.042 < 0.10 and 0.042<0.08

Therefore,the largest level of confidence for which the confidence interval will NOT contain 11 is 98% confidence level.

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Find the area of the circle with a circumference of
Anton [14]

Answer:

706.50 mm^2

Step-by-step explanation:

The formula to find the circumference is 2πr (r : radius)

if the circumference is 30π the the radius is = 15

to find the area we use the following formula :

π*r^2

r = 15 and π is given as 3.14

(3.14)*15^2 = 706.50 mm^2

8 0
2 years ago
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
Could someone help me please
Rina8888 [55]

Answer:

the answer should be 30

Step-by-step explanation:

since you can divided 15 by 5 to get 3, you can multiply 10 x 3 to get 30

8 0
3 years ago
Circle theorem - easy work
makvit [3.9K]
Angle in a semi circle is 90°
<a = 180-90-<b = 90-28 = 62°

opposite angles of a cyclic quadrilateral sums up to 180°
<b + <a = 180
<b = 180-62
<b = 118°
8 0
3 years ago
HELP AGAIN PLEASE I TRIED IT ON MY OWN AND FAILED :(
Alex787 [66]

Answer:  ∠ J = 62° , ∠ K = 59° , ∠ L = 59°

<u>Step-by-step explanation:</u>

It is given that it is an Isosceles Triangle, where L J ≅ K J

It follows that ∠ K ≅ ∠ L

         ⇒ 5x + 24 = 4x + 31

         ⇒   x + 24 =         31

         ⇒   x         =           7

Input the x-value into either equation to solve for  ∠ K &  ∠ L:

∠ K = 5x + 24

      = 5(7) + 24

      = 35 + 24

       = 59

∠ K ≅ ∠ L  ⇒   ∠ L = 59

Next, find the value of  ∠ J:

∠ J +  ∠ K +  ∠ L = 180    Triangle Sum Theorem

∠ J  +   59  +  59 = 180

∠ J          + 118     = 180

∠ J                       = 62

8 0
3 years ago
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