Statistics help? Confidence Interval?

So I was sick and missed a class and the next week we were doing this together. I kind of understand the concept, but Im not sure what I need to do first to solve this now that I'm studying on my own. So no more than 2 points would be P1-2? And since its a sample you have to find Phat which would mean 4+12/67 (the population?) So that would equal .239, but after finding Phat what do you need to do? Can someone explain to me  where we got . 2.576? Is this Z?

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  • Alan
    Lv 7
    3 weeks ago

    99% confidence

    interval goes from 0.5 % to 99.5 %

    which gives you 99% in the middle

    so look up P(z<Z) =0.995 in a z-table

    Using this table

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

    Look for closest value to 0.995 inside the table

    they are the following:

    p(z< 2.57) = 0.99492

    P(z<2.58) =0.99506

    Now, interpolate

    =2.57 + (0.995 - 0.99492)*0.01/ (0.99506-0.99492)

    =2.575714286

    Now , if you rou nd to 3 places after the decimal place

    = 2.576

    so the P(z< 2.576) = approx. 0.995 (which is the upper limite of 99% confidence level)

    go further

    so yes,

    p_hat = (4+12)/67

    p_hat = 16/67 = 0.23880597

    so for part. a

    just use the formula

    CI_low = p_hat + z_critical* standard error

    CI_High =p_hat + z_critical*standard error

    z_critical = 2.576

    standard error = sqrt(p *(1-p)/n) = sqrt((16/67)( 51/67)/67))

    standard error = 0.052087423

    https://stattrek.com/estimation/standard-error.asp...

    σp = sqrt [ P(1 - P) / n ]

    CI_low = (16/67) -2.576*0.052087423

    CI_low = 0.104628769

    CI_high = (16/67) +2.576*0.052087423

    CI_High = 0.372983172

    so interval is approx. ( 0.1046 , 0.3730)

    Part B

    Alpha = 0.10

    but the question is less than

    so you are looking for p =0.10 since it is one-tailed

    P(z< Z_critcal) = 0.10

    P(z< -1.29) = 0.09853

    P(z< -1.28) = 0.10027

    if you interpolate , you get about this

    z = -1.281551566

    I will use z = -1.2816

    1 of 3 = 1/3

    p_hat_test = 1/3

    z_test = (measured -1/3)/ standard error

    p(dissatisfied test) = 12/67

    so if you are very dissatified are you still dissatified

    then

    P_test = (12+4)/67 = 16/67 again

    z_test (just dissatified) = ((12/67) - (1/3) /0.052087423

    z_test = -2.960961504

    z_test = approx. -2.961

    since z_test is less z_critical

    -2.961 < -1.2816

    or if you use

    p_test = 16/67 (combining dissatfied and very dissatified )

    z_test(combined 1 and 2) = ( (16/67) -(1/3) ) / 0.052087423

    z_test (combined 1 and 2) = -1.814782857

    so -1.815 < -1.2816

    so in either case ,

    so you can reject the NULL Hypothesis and

    support the claim that less than 1 of 3 are dissatified

    because z_test< z_critical

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