# 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?

### 1 Answer

- AlanLv 73 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