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rosijanka [135]
3 years ago
14

A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standa

rd deviation were calculated. Based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 4% of 70%.
Is the confidence interval at 90%, 95%, or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.
Mathematics
1 answer:
swat323 years ago
4 0
All our answers lie in the above statement.

Confidence Level:
The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%. 

Margin of Error:
The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .

Confidence Interval:
Lower Limit = 70% - 4% = 66%
Upper Limit = 70% + 4% = 74%

Interpretation:
The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%. 
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Answer:

a. Weekly Gross pay:

= 9.50 * 20

= $190

b. Federal tax withheld:

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= $19

c. FICA Tax withheld:

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= $10.74

d. State tax withheld:

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e. Weekly net pay

= Gross pay - taxes

= 190 - 19 - 10.74 - 9.50

= $150.76

f. Percentage withheld for taxes:

= (19 + 10.74 + 9.50) / 190 * 100%

= 20.7%

8 0
3 years ago
0.25r+0.6s. What is the verbal expression
drek231 [11]
0.25 times r added to 0.6 times 6
3 0
3 years ago
Read 2 more answers
Suppose 232subjects are treated with a drug that is used to treat pain and 50of them developed nausea. Use a 0.01significance le
Margarita [4]

Answer:

A

   The  correct option is B

B

   t =  0.6093

C

 p-value  =  0.27116

D

The  correct option is  D

Step-by-step explanation:

From the question we are told that

    The  sample size is  n  =  232

    The  number that developed  nausea  is X =  50

    The population proportion is  p  =  0.20  

 

The  null hypothesis is   H_o : p  =  0.20

The  alternative hypothesis is  H_a :  p > 0.20

Generally the sample proportion is mathematically represented as

     \r p  =  \frac{50}{232}

     \r p  =  0.216

Generally the test statistics is mathematically represented as

 =>           t =  \frac{\r p  -  p }{ \sqrt{ \frac{p(1- p )}{n} } }

=>           t =  \frac{ 0.216 - 0.20 }{ \sqrt{ \frac{ 0.20 (1- 0.20 )}{ 232} } }

=>        t =  0.6093

The  p-value obtained from the z-table is

       p-value  =  P(Z >  0.6093) =  0.27116

  Given that the  p-value >  \alpha  then we fail to reject the null hypothesis

5 0
3 years ago
Brian and Bob visit a ski resort. Bob buys 2 full passes and 3 restricted passes. The total cost of his passes is £130.
Jobisdone [24]

Answer:

2x+3y=130

Step-by-step explanation:

Let x=cost of full pass

Let y=cost of restricted pass

Brian and Bob bought 2 full and 3 restricted passes.

Have a good day!

6 0
3 years ago
MAJOR HELP!
Feliz [49]

Answer:

The radius of the cone is 21 units

The volume of the cone is 9702 units³

Step-by-step explanation:

* Lets revise the area of the circle and the volume of the cone

- The base if the cone is a circle

- The area of the circle is πr², where r is the radius of the circle

∵ The area of the base of the cone = 1386 units²

∵ The area of the base = πr²

∵ π = 22/7

∴ 1386 = 22/7 (r²) ⇒ multiply each side by 7

∴ 9702 = 22 r² ⇒ divide both sides by 22

∴ 441 = r² ⇒ take √ for both sides

∴ 21 = r

* The radius of the cone is 21 units

- The volume of the cone = 1/3 πr²h , where h is the height of the cone

∵ The cone's height is equal to the radius of its base

∴ h = r

∵ The volume of the cone = 1/3 πr²h

∴ The volume of the cone = 1/3 πr²(r) = 1/3 πr³

∵ r = 21 units

∴ The volume of the cone = 1/3 (22/7) (21)³

∴ The volume of the cone = 9702 unit³

* The volume of the cone is 9702 units³

3 0
4 years ago
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