A standard medication usually cures about 75 of patients wit
A standard medication usually cures about 75% of patients with certain disease. If a new medication cured 310 of the first 400 patients on whom it was tried, can it be concluded that the new medication is better?Test the hypothesis at 5% significance level.
Since the observed value 310 is not greater than c (=314), we do not reject the hypothesis.This indicates that the results obtained so far do not establish the superiority.
Since the observed value 310 is greater than c (=300), we can reject the hypothesis.This indicates that the results obtained so far establish the superiority.
Since the observed value 310 is greater than c (=200), we can reject the hypothesis.This indicates that the results obtained so far establish the superiority.
Since the observed value 310 is not greater than c (= 514), we do not reject the hypothesis.This indicates that the results obtained so far do not establish the superiority.
| Since the observed value 310 is not greater than c (=314), we do not reject the hypothesis.This indicates that the results obtained so far do not establish the superiority. | ||
| Since the observed value 310 is greater than c (=300), we can reject the hypothesis.This indicates that the results obtained so far establish the superiority. | ||
| Since the observed value 310 is greater than c (=200), we can reject the hypothesis.This indicates that the results obtained so far establish the superiority. | ||
| Since the observed value 310 is not greater than c (= 514), we do not reject the hypothesis.This indicates that the results obtained so far do not establish the superiority. |
Solution
Z-Test For Proportion
Set Up Hypothesis
Under The Null Hypothesis H0:P=0.75
Under The Alternate Hypothesis H1: P>0.75
Test Statistic
No. Of Success chances Observed (x)=310
Number of objects in a sample provided(n)=400
No. Of Success Rate ( P )= x/n = 0.78
Success Probability ( Po )=0.75
Failure Probability ( Qo) = 0.25
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.775-0.75/(Sqrt(0.1875)/400)
Zo =1.15
| Zo | =1.15
Critical Value
The Value of |Z ?| at LOS 0.05% is 1.64
We got |Zo| =1.155 & | Z ? | =1.64
Make Decision
Hence Value of |Zo | < | Z ? | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 1.1547 ) = 0.12411
Hence Value of P0.05 < 0.12411,Here We Do not Reject Ho
Since the observed value 310 is greater than c (=300), we can reject the hypothesis.This indicates that the results obtained so far establish the superiority.
