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How do you find the P value of a hypothesis test?

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its the area of the tails ....
determine the z or t score using the Ha as the cutoff stat for Ho the P-value is just the area left in the tails
This is my particular problem: A survey of 1910 people who took trips revealed that 136 of them included a visit to a theme park. Based on those survery results, a management consultant claims that less than 8% of trips include a theme park visit. Test this claim using the alpha= .01 significance level. Assuming Ho, The z score is? The critical z score for declaring significance is? The P Value of the Hypothesis test is_____?

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Other answers:

are you using a ti83, stat program, or just tables in the book?
I found the z score to be -1.4168...then the critical z score using norm.s.inv(.99) on excel to be 2.32...but I do nt know how to find the p value...would I need to find the z score for .08 and use that value into norm.s.dist(z value, cumulative)? that is what I think to do...but I don't know how to find the z value with P (x<.08)
we have been using I am so glad I found this site, I have been searching for a free help site and had no luck
1910*.08 = 153 use 153 is your test value
since there is no standard deviation given, should we do a t stat, or a z stat? sample size seems well over 30 to me
what is a t stat?
I believe mean z score?
its similar to a zscore stat, but uses a t distribution as opposed to a normal distribution .... its used for unknown standard deviations
@ViennaJade , Excel gives pvalue of .078 for z-score of -1.4168 "NORMSDIST(-1.4168)"
hmmm dumbcow let me try that
admiter I found the standard deviation by hand
136/1910 = .0712 as a "mean" this is less then 8% of the sample size already so i don think the score of 8% will be negative
I just checked dumbcow's answer and it was right
now I'm not sure I remember why his answer is right though...
\[\Large\frac{136}{1910}\pm Z_{.01}\left( \frac{\frac{136}{1910}\frac{1774}{1910}}{\sqrt{1910}} \right)\].01 in one tail is .02 in both tails,relates to a confidence interval of 98%, z=2.33 a confidence interval of 98% is \[\Large .0712\pm 2.33\frac{(.0712)(.9288)}{\sqrt{1910}}\] cows answer was correct, thats nice :)
opps, forgot to sqrt the whole thing
why are you finding the confidence interval? I'm confused
so an interval of .057 to infinity .... just trying to recall some stuff. how would you determine the test statistic for less then 8%?
I would want to use the z score formula , plug in 153/1910-.08 all over the standrd dev of .006208
and then use excel =norm.s.dist(the z score)
.08 - 153/1910 of course the only differnce is a sign
amistre thank you so much for helping me...and you get paid for this? this problem is a 1-proportion z test right? \[z = \frac{.08 - \frac{136}{1910}}{\sqrt{\frac{136}{1910}(1-\frac{136}{1910})}}\sqrt{1910}\]
dumbcow I have no idea what that equation is...we were taught that z score of p hat= p hat-Ho/stdev
how are you finding std dev?
I don't think it's 1-proportion because they want less than .08
stdev of p hat= sqrt(p*q/n)
p is .08, q is .92
n is 1910
oh ok, well in that case you already did everything and didn't need any help :)
I still didn't get how you used the z score I found and used that to find the p(x<.08)? sorry..
|dw:1370971511146:dw| the p-value is the probability of being "left" of z-score on the bell curve the excel function "normSDist" will tell you the area "left" of a given z-score
if p-value < .01 , reject H0 otherwise Fail to reject H0
I think I get it. Thank you so much Amistre and dumbcow! :))

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