Hello, Can anyone please explain how to find the critical value? ?

Hello, Can anyone please explain how to find the critical value? For example z0.01= 1.28, how do you find the 1.28? The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. They would like the estimate to have a maximum error of 0.13 gallons. A previous study found that for an average family the standard deviation is 1.3 gallons and the mean is 18.5 gallons per day. If they are using a 98% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer.

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  • Alan
    Lv 7
    1 month ago
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    Critical value for two-sided test or hypothesis test where you are 

    checking whether the mean is different  You look the probability P(z< Z) = 1- alpha/2  for a  one sided test,   where H1 is > or  < 

    then ,, you look up 

    P(z< Z)  = 1 -alpha   for a right side test 

    p(z<Z ) = alpha for  left tail  

    so if alpha = 0.10 and your test is one-sided 

    you look up 

    P(z<Z)  = 1- 0.10  = 0.90  

    so in Excel =normsinv(0.90) = 1.281551566 to the nearest hundredth 

    Z =1.28  

    or looking up in  a Z-table 

    P(z< 1.28) =  0.89973

    P(z< 1.29) = .0.90147 

    From here some courses/professor would expect you to 

    interpolate . 

    Other course/professor tell you to just use the closest 

    value to 0.90  

    so Z = 1.28 is closer to 0.90  than 

    Z = 1.29  

    since  0.9 - 0.89973   =     0.00027     

    and 0.90147 - 0.90 =    0.00147

    so 0.00027   

    if you interpolate, you should come close to matching  

    the same value from  a normal calculator or excel normal function. 

    with alpha = 0.10 and a two-sided test or a confidence 

    interval, you would look up 

    P(z<Z) = 1-  0.10/2 = 0.95    

    so in this case , 1.28  would be wrong 

    ==== In your current problem  

    You have a confidence range which is two-sided.

    If you are just given a confidence range  

    alpha = 1 - confidence range 

    P(  z<Z ) =   1 -(1 -confidence range)/2  = 1 - (1-0.98)/2  = 1 - 0.02/2 

    P(z<Z ) = 0.99 

    and this makes sense  because 0.99-0.01 = 0.98  

    so you have a 98 Percent range. 

    so look up this value using an inverse normal function 

    like =normsinv(0.99) in  Excel  

    or find the two closest value in z-table 

    P(z< 2.32) = .0.98983 

    P(z< 2.33) = 0.99010 

    so 2.33 is the closest value  (0.00010 above, versus 0.00017 below) 

    If you interpolate 

    2.32  + (0.99- 0.98983)*(0.01)/ (0.99010-0.98983) 

    =2.3263     

    so now this is your problem find out what you course, teacher or

    professor wants you to use.  

    Some course use the closest value , so they would 

    expect you to just  use 2.33

    while others may expect to  

    interpolate only when it is off by a certain amount 

    while others course may expect you interpolate all the time. 

    CI_Low = mean - Z_critical*standard deviation/sqrt(N) 

    CI_High = mean - Z_critical*standard deviation/sqrt(N) 

    so the error on both sides is 

    SE = Z_critical*standard deviation/sqrt(N)  

    To find max error ,  0.13 gallons 

    SE =  0.13  gallons 

    Z_critical = 2.3263 (interpolated value ) or 2.32 (closest value ) 

    Standard deviation = 1.3  

    0.13 =  2.3263*1.3/sqrt(N) 

    multiply both sides by sqrt(N) 

    sqrt(N)*0.13  = 2.3263*1.3 

    divide both sides by 0.13 

    sqrt(N) = 2.3263*1.3/0.13

    square both sides 

    N =   (2.3263*1.3/0.13)^2

    N =  541.167169

    N = 542  (rounded up) 

    If you have used the closest value 

    N= (2.33*1.3/0.13)^2 = 542.89

    N = 543  (using closest Z-value) 

    N = 542 (using interpolated Z-value)  

    So again, you have to learn what is expected in your course  

    as to when or if you should interpolate 

    or whether you can just use 

    Excel or calculator normal functions. 

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