A new drug has a potency advertised at 75%. Given the following sample data: n=2

Question

A new drug has a potency advertised at 75%. Given the following sample data: n=25, x=74.6%, s=1% Check manufacture's claim of 75% potency at alpha=.05 level (one tail). Make 90% confidence interval. If the desired error of estimation is less than 0.1% with 99% confidence, what the sample size be? An experiment was conducted to compare how taking the elevator versus taking stairs to the third floor of Mackay affected the heat rates of 4 students. Does the below provide sufficient evidence that taking the stairs elevates the heat rate? Test alpha=.05 level with a two tail test. IN the year 2001, the average GPA of College Graduates was 3.1. A recent survey 1000 students determined an average GPA of 3.3 with a standard deviation of 3. Do a two tailed test at alpha=.05 to determine if there is grade inflation since 2001. Make a 95% confidence interval. If the desired error of estimation is less than 0.14 with 98% confidence, what must the sample size be?

Explanation / Answer

1.

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   75  
Ha:    u   <   75  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    24          
tcrit =    -   1.71088208      
              
Getting the test statistic, as              
              
X = sample mean =    74.6          
uo = hypothesized mean =    75          
n = sample size =    25          
s = standard deviation =    1          
              
Thus, t = (X - uo) * sqrt(n) / s =    -2          
              
Also, the p value is              
              
p =    0.028469925          
              
As |t| > 1.711, we   REJECT THE NULL HYPOTHESIS.  

Hence, there is significant evidence at 0.05 level that the true mean percent potency is less than 75%. [CONCLUSION]

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b)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    74.6          
t(alpha/2) = critical t for the confidence interval =    1.71088208          
s = sample standard deviation =    1          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    0.342176416          
Lower bound =    74.25782358          
Upper bound =    74.94217642          
              
Thus, the confidence interval is              
              
(   74.25782358   ,   74.94217642   ) [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    1  
E = margin of error =    0.1  
      
Thus,      
      
n =    663.4896601  
      
Rounding up,      
      
n =    664   [ANSWER]

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