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Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

4 years ago

What is (-2) (3 4/7)

Rina8888 [55]

Answer:

option 1

Step-by-step explanation:

(-2) (25/7)

-50/7 = -7 1/7

8 0

2 years ago

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Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0

Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

$h(x)=\dfrac{1}{3}|x-6|+4$

Put the values of x to the equation of the function h(x):

$x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6$

6 0

3 years ago

Solve by Substitution Show Steps −2x + 3y + 5z = −21 −4z = 20 6x − 3y = 0

Anarel [89]

−2x + 3y + 5z = −21

−4z = 20

6x − 3y = 0

do -4z=20 first

divide both sides by -4 to get z by itself

-4z/-4=20/-4

z=-5

Use z=-5 into −2x + 3y + 5z = −21

-2x+3y+5(-5)=-21

-2x+3y-25=-21

move -25 to the other side

sign changes from -25 to +25

-2x+3y-25+25=-21+25

-2x+3y=4

6x-3y=0

find x by eliminating y

Add the equations together

-2x+6x+3y+(-3y)=4+0

-2x+6x+3y-3y=4

4x=4

Divide by 4 for both sides

4x/4=4/4

x=1

Use x=1 into 6x − 3y = 0

6(1)-3y=0

6-3y=0

Move 6 to the other side

6-6-3y=0-6

-3y=-6

Divide both sides by -3

-3y/-3=-6/-3

y=2

Answer:

(1, 2, -5)

5 0

4 years ago

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Please help me thank you

nexus9112 [7]

It is 9 because 11 minus 2 is equal to 9.

3 0

3 years ago

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