2007 divided by j = 223

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% 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 average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 96% confidence interval for the population mean, $\mu$ is ;

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0

3 years ago

Multiply the following rational expressions and simplify the result

GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$ = $\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}$

= $\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}$

= $\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}$

= $\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}$

3 0

3 years ago

Sam complete 3 math problems in 8 minutes. How long will it take to do 15 problems?

pochemuha

Answer:

40 minutes

Step-by-step explanation:

Math Problems 3 15

---------------------- ----- = ------

Minutes 8 x

3x = 15 * 8

x = 15 * 8 / 3

x = 120 / 3

x = 40

40 Minutes

8 0

3 years ago

What is a equivalent fraction for 3/9?

kherson [118]

1/3 = 2/6 = 3/9 = 4/12 = 5/15 = 6/18 = 7/21 = 8/24 = 9/27 = 10/30 = 11/33 = 12/36 = 13/39 = 14/42 = 15/45 = 16/48 = 17/51 = 18/54 = 19/57 = 20/60
You can use any, these are just the smallest to the biggest but any are fine but I would use the first one. Hope I helped. Equivalent is what 3/9 equals to.

5 0

3 years ago

I need Help plz...!!!!

kirill [66]

Answer:

B.21.3 is the correct answer

Step-by-step explanation:

1/2*x+1/4*x=16

x/2+x/4=16

2x+x/4=16

3x/4=16

3x=16*4

3x=64

x=21.3

Hope it is helpful for you

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6 0

3 years ago

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