Select all the expressions with a value of less than one.

The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh

MAXImum [283]

Answer:

a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`= 6,700 hours

Sample standard deviation=s= 600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282

Z(0.05)=1.645

Z(0.025)=1.960

t(0.01)(49)= 1.299

t(0.05)= 1.677

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0 for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test we reject H0 as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ± z∝/2 (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0

3 years ago

"How are straight angles different from right angles?" Is this a statement? <br> yes or no

Arturiano [62]

No, this is a question.

7 0

3 years ago

What is the domain for each function below? Choose the correct inequality symbol from the drop down menu and fill in the blank w

erik [133]

Answer:

$x \le 3$

$x \ge 0$

Step-by-step explanation:

The functions, missing from the question are:

$f(x) = \sqrt{3 - x$

$f(x) = \sqrt{3 x$

Required

Determine the domain

To get the domain, means we want to get the possible values of x.

(a)

$f(x) = \sqrt{3 - x$

Equate 3 - x to 0

$3 - x = 0$

Solve for x

$-x = -3$

$x = 3$

From the calculated value of x above;

When x exceeds 3; f(x) = undefined

So: the domain is:

$x \le 3$

(b)

$f(x) = \sqrt{3x$

Equate 3x to 0

$3x = 0$

Solve for x

$x = 0$

From the calculated value of x above;

When x falls below 0; f(x) = undefined

So: the domain is:

$x \ge 0$

4 0

2 years ago

Jose walks at a rate of 4 miles per hour and runs at a rate of 8 miles per hour. Each week, his exercise program requires him to

Nastasia [14]

Answer:

Here we can use the relationship:

Distance = time*speed.

When Jose walks, his speed is 4 mph.

Then if he walks for X hours, the distance that he will travel is:

D = 4mph*X

When Jose runs, his speed is 8mph.

Then if he runs for T hours, the distance that he will travel is:

D´ = 8mph*Y

And we know that he travels in total 20 miles, then we must have that:

D + D´= 20mi

This leads to:

4mph*X + 8mph*Y = 20mi

Where X is the time that he walked, and Y is the time that he runed.

Then the equation that represents the different amounts of times that Jose runs and walks is:

4mph*X + 8mph*Y = 20mi

Where we can not really find the solutions for Y and X, because there is only one equation and two variables.

7 0

2 years ago

A+ bakery uses vanilla extract in the baking of their cookies. a small bottle of vanilla extract has 16 ounces in it. if each ba

yaroslaw [1]

Answer:

24 times

Step-by-step explanation:

Given : A small bottle of vanilla extract has 16 ounces in it.

To Find : if each batch of cookies baked requires 2/3 of an ounce of vanilla how many times can the bakery use the bottle before it is empty?

Solution :

No. of batch of cookies requires 2/3 ounce of vanilla = 1

No, of batch of cookies requires 1 ounce of vanilla = $\frac{1}{\frac{2}{3}}={\frac{3}{2}}$

No. of batches require 16 ounces of vanilla = $\frac{3}{2} * 16$

= 24

Thus, bakery can use 24 times the bottle before it is empty.

8 0

3 years ago

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