Anonymous
Anonymous asked in Education & ReferenceHomework Help · 4 weeks ago

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. 

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  • Alan
    Lv 7
    4 weeks ago
    Best 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

    • Alan
      Lv 7
      4 weeks agoReport

      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

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