answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Musya8
1 month ago
12

One division of a large defense contractor manufactures telecommunication equipment for the military. This division reports that

12% of non-electrical components are reworked. Management wants to determine if this percentage is the same as the percentage rework for the company’s electrical components. The Quality Control Department plans to check a random sample of the over 10,000 electrical components manufactured across all divisions. The 95% confidence interval based on this data is .0758 to .1339. Should management conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components?
Mathematics
2 answers:
babunello [11.8K]1 month ago
8 0

Answer:

It is not possible to claim that the rework percentage for electrical components is less than the 12% for non-electrical components.

Step-by-step explanation:

The 95% confidence interval indicates that the true mean for the rework percentage of electrical components is between 0.0758 (7.58%) and 0.1339 (13.39%), maintaining a 95% confidence level.

The reported figure of 12% falls within this range, suggesting the actual mean could indeed be 12%. This means we cannot dismiss the possibility that the rework percentage is 12%.

<pConsequently, we cannot infer that the proportion of rework for electrical components is below the 12% recorded for non-electrical components.
Leona [12.6K]1 month ago
6 0

Answer:

Under a 95% confidence level, management should refrain from concluding that the rework percentage for electrical components is below the 12% observed for non-electrical components. This is because the upper limit of the confidence interval, which stands at 0.1339, exceeds 0.12.

Step-by-step explanation:

To arrive at a conclusion, it is necessary to analyze the confidence interval.

Should management assert that the rework percentage for electrical components is less than the 12% for non-electrical components?

Is the upper limit of the confidence interval below 12% = 0.12?

If this is true, they could claim that the percentage is lower.

If not, they cannot make that claim.

Confidence interval:

0.0758 to 0.1339

Notably, 0.1339 is greater than 0.12.

Therefore,

At the 95% confidence interval, management should not assert that the rework percentage for electrical components is below 12%, as the upper limit of the confidence interval, 0.1339, surpasses 0.12.

You might be interested in
What is the domain of the function graphed below?
zzz [12365]
The function is applicable within the segments of x:
(-∞, -1) and [-1, 7), meaning it is valid for x < 7.

Importantly,
the function cannot be evaluated at x = -1 in the left part of the linear graph, while it is valid at x = -1 in the right segment of the same line. Additionally, the function is not defined at x = 7 or any value above it.

Conclusion: x < 7.
5 0
1 month ago
Read 2 more answers
Mara ran 3km north and then 4km east. She will finish her run by running directly home. What was the total distance of her run?
tester [12383]
To find the hypotenuse of a right triangle with sides measuring 3 and 4, we first need to use the Pythagorean theorem and then add that distance to 3 and 4.

By applying the theorem, the square of the hypotenuse equals the sum of the squares of the sides...

d^2=3^2+4^2

d^2=9+16

d^2=25

d=√25

d=5

Thus, the total distance for her run is 5+4+3=12 km
5 0
19 days ago
Read 2 more answers
a jewelry is designing a crown. the crown must have at least 45 gems using only emeralds, rubies, and diamonds. the number of ru
Inessa [12570]
I'm quite terrible at this aspect, so I'm merely making a guess.
X serves as the variable while x denotes the multiplication symbol.
X + (X x 3 - 1) + (X x 3 - 1) - 4) <span>≥ 45 is the outcome I arrived at.
The solution to that inequality is
X </span><span>≥ 7, I suppose.</span>
5 0
20 days ago
A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
AnnZ [12381]

Response:

a. As student debt rises, current investment diminishes.

b. Y= 68778.2406 - 1.9112X

For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.

c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step breakdown:

Hello!

Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:

Y: Current investment by an individual who graduated from college five years prior.

X: Total debt of an individual upon graduating five years ago.

a)

To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.

The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.

Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develop this equation, estimates for alpha and beta are required:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

b=\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Averages:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation becomes:

Y= 68778.2406 - 1.9112X

For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.

c)

To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)

Using t=  b - β  =  -1.91 - 0  = -31.83

                 Sb         0.06

The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001

Since this P-value is underneath the significance level, we reject the null hypothesis.

In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.

The P-value remains: 0.0001

Using this method, we similarly reject the null hypothesis.F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

In conclusion, at a significance level of 1%, there exists a linear relationship linking current investment to college debt.

The accurate statement is:

There exists a significant linear association between college debt and current investment since the P-value is less than 0.1.

d)

To forecast the value of Y when X is set, it is essential to substitute X in the estimated regression equation.

Y/$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The anticipated investment for someone with a college debt of $5000 is $59222.2406.

e)

To determine the proportion of variation in the dependent variable that the independent variable accounts for, the coefficient of determination R² must be calculated.

R²= 0.9818

R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }

R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }

This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X

I trust this is beneficial!

5 0
12 days ago
Which table shows a function that is decreasing only over the interval (–1, ∞)? A 2-column table with 6 rows. The first column i
Inessa [12570]
B. f(x) ≤ 0 over the interval [0, 2]. D. f(x) > 0 over the interval (–2, 0). E. f(x) ≥ 0 over the interval [2, ).
8 0
1 month ago
Other questions:
  • A set of notations (SSS, SAS, ASA and RHS) is used to describe/prove
    8·1 answer
  • Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)
    8·1 answer
  • Wang Yong owes the bank $8500. To repay the debt, he paid a fixed amount back to the bank each month. After 12 months, his remai
    13·1 answer
  • John's average for making free throws in a basketball game is .80. In a one-and-one free throw situation (where he shoots a seco
    6·2 answers
  • George is 1.94 meters tall and wants to find the height of a tree in his yard. He started at the base of the tree and walked 10.
    12·1 answer
  • Complete the equation of the line through (2,-2)(2,−2)left parenthesis, 2, comma, minus, 2, right parenthesis and (4,1)(4,1)left
    15·1 answer
  • An element with mass 730 grams decays by 28.8% per minute. How much of the element is remaining after 10 minutes, to the nearest
    10·2 answers
  • Martene got a small aquarium for her birthday. The aquarium is a right rectangular prism 18.5\,\text{cm}18.5cm18, point, 5, star
    11·1 answer
  • Smoothie Stand at Utah’s Wasatch County’s Demolition Derby sells bananas. If John bought 55 lbs. of bananas at $.24 per pound ex
    12·2 answers
  • A group of dragons and sheep are randomly divided into 2 equal rows. Each animal in one row is directly opposite an animal in th
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!