Question 1191869: 50 people were attacked by a disease and 30 were survived. If the survival rate is 70%.Test the claim at 5% level of significance.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 30/50 = 60%.
z-score = (x - m) / s
z is the z-score
x is the raw score
m is the mean you are comparing against.
s is the standard error.
s = sqrt(70 * 30)/50) = 6.4807
z = (x - m)/s = (60 - 70) / 6.4807 = -1.54
critical z-score at 5% two sided level of significance is plus or minus 1.96.
-1.96 is more negative than -1.54.
there is not enough evidence to determine that the 70% can be rejected because the results are not significant (not greater than the critical z-score).
if you did this with ratios and not percents, the results would be the same.
s = sqrt(.7*.3/50) = .064807
z = (.6 - .7) / s becomes z = -.1 / .064807 = -1.54.
same z-score.
same result.
here's a reference.
https://online.stat.psu.edu/statprogram/reviews/statistical-concepts/proportions
even if you had done a one tailed test, the critical z-score would have been plus or minus 1.645.
-1.54 is still not greater than -1.645, so the results are still not significant.
this means not enough evidence to overturn the presumed ratio of .7.
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