# 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.

### 1 Answer

- AlanLv 71 month agoFavorite Answer
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.