# I need statistics 1602 help. I just want to know how to work them out. ?

I don't really understand how my teacher teaches and i have the answers to the worksheet but I want to know how to be able to get them not just the answers.

### 1 Answer

- AlanLv 72 months agoBest answer
I showed you 1-5 , I didn't round my answer

thinking you can do your own rounding.

Ask question if you don't understand and try

the rest of the questions.

Formulas to know,

For a proportion ,

standard error = sqrt(p * (1-p)/n)

p_low= p - z_critical*standard error

p_high= p + z_critical*standard error

for sample mean cases,

CI_low = mean - z_critical*standard deviation/sqrt(N)

CI_High = mean + z_critica*standard deviaton/sqrt(N)

For all z-table information, I used this

table

https://www.math.arizona.edu/~rsims/ma464/standard...

so what is z_critical

so for XX % confidence range

it goes from

(100-XX)/2 to 100 - (100-XX)/2

so a 95 % confidence range

goes from 2.5 % to 97.5 %

then,

P(z < z_Critical) = 0.975

P(z< 1.96) = 0.975

z_critical = 1.96

so 99% confidence range

it goes from 0.5 % to 99.5 %

so z_critical Z such that

p(z<Z) = 0.995

looking up the two closest value in a z-table gives

P(z< 2.58) = 0.99492

P(z< 2.59) = 0.99506

Interpolating

=2.58 + (0.995-0.99492)*0.01/ (0.99506-0.99492)

Z_critical = approx. 2.5758

so 90% confidence range ,

it goes from 5 % to 95%

so z_critica is z such

P(z< 1.64) = .94950

P(z< 1.65) = .95053

since it is almost exactly in the center

P(z< 1.645) = approx. 0.95

z_critical = 1.645

so for 95% confidence range , z_critical = 1.96

for a 99% confidence range , z_critical = 2.5758

for a 90 % confidence range , z_critical = 1.645

so for all the question with these defined

ranges you can use these values

1.

Use the p formulas

For a proportion ,

standard error = sqrt(p * (1-p)/n)

p_low= p - z_critical*standard error

p_high= p + z_critical*standard error

so p = 41/500 = 82/1000 = 0.082

1 -p = 0.918

z_critical = 1.96

n = 500

just plug into the formula

p_Low = 0.082 - 1.96* sqrt( 0.082*0.918/sqrt(500) )

p_high =0.082 + 1.96*sqrt( 0.082*0.918/sqrt(500) )

p_low = 0.057950872

p_high = 0.106049128

2.

since the formula are

CI_low = mean - z_critical*standard deviation/sqrt(N)

CI_High = mean + z_critica*standard deviaton/sqrt(N)

the CI ranges from mean +/- z_critical*standard deviation/sqrt(N)

so it is within

+/- z_critical*standard deviation/sqrt(N)

since it is a 99 % confidence range

z_critical = 2.5758 approx.

so you have

0.2 = z_critical*standard deviatiion/sqrt(N)

0.2 = 2.5758 * 1 /sqrt(N)

multiply both sides by the sqrt(N)

0.2*sqrt(N) = 2.5758

divide both sides by 0.2

sqrt(N) = 2.5758/0.2 = 128.79

square both sides

N = 128.79^2 = 16586.8641

since n must be an integer and n >= 16586.8641

N = 16587

3. Just go back the formulas

so again a 99 % confidence range , so z_critical = 2.5758 again

N = 40

mean = 20

standard deviation = 1.5

some text books may think 40 is too small of a value for use z-table

then you would use t_critical(DF = 39 , alpha = 0.5 (0.995 value) )

CI_low = mean - z_critical*standard deviation/sqrt(N)

CI_High = mean + z_critica*standard deviaton/sqrt(N)

CI_low = 20 - 2.5758*1.5/sqrt( 40)= 19.38909539

CI_high =20 + 2.5757*1.5/sqrt(40) = 20.61090461

b. Is it possible , yes

but it is a very low probability which is less than 0.5 % just based on the number we found

in part a.

4. again we want a value BE LESS THAN

THE RANGE

= +/- z_critical*standard deviatiion/sqrt(N)

standard deviation - 0.068

z_critical (since 99% again) = 2.5758

allowable range = 0.025

0.025 = 2.5758*0.068/sqrt(N)

multiply both sides by the sqrt(N)

0.025*sqrt(N) = 2.5758*0.068

divide both sides by 0.025

sqrt(N) = 2.5758*0.068/0.025 = 7.006176

square both sides

N = 7.006176^2 = 49.08650214

but since n >= 49.086 and it must be an integer

N = 50

5.

Right back to the same formulas

mean = 1.23

standard deviation = 0.65

N = 1120

z_critical= 1.645

CI_low = 1.23 - 1.645*0.65/sqrt( 1120)

CI_Low = 1.198050045

CI_High = 1.23 + 1.645*0.65/sqrt(1120)

CI_High = 1.261949955

For question 6,

use this formula to determine z-values

:::::: z_test = (x-mean)/ (standard deviation/sqrt(N) )

:::: z_test = (4995-5000)/ (16/sqrt(64)) = -5/2 =-2.5

:::: then , look up P(z< -2.5) = .00621

:::: since , 0.00621 < 0.01 , you can support claim

and reject NULL Hypothesis