A Researchers wanted to determine if one of the effects of a new drug is a decre

Question

A Researchers wanted to determine if one of the effects of a new drug is a decrease in hemoglobin levels. So they test the levels in a rat's hemoglobin before the drug and then after the drug. The hemoglobin (in grams per deciliter) at three days after the drug and the hemoglobin before the rat took the drug. This was done for 14 randomly selected rats. Use the MINITAB output to help you determine if the evidence suggests that the drug decreases hemoglobin levels at the alpha = 0.05 significance. (The difference is after drug - before drug.) The probability plot and boxplot of the differences in Hemoglobin levels. Use the above information to answer questions 32 - 35. 32. State the null and alternative hypothesis (mu_d is the difference of means for matched pairs.) H_0: mu_d = 0 vs H_a: mu_d notequalto 0 H_0: mu_d = 0 vs H_a: mu_d 0 Cannot be done. What is the critical value? State whether you reject or do not reject H_0. Reject Do not reject Cannot be done. Construct a 99% confidence interval, 0 plusminus 0.798 - .875 plusminus 0.933 0 plusminus 0.933 Cannot be done

Explanation / Answer

32.

Formulating the null and alternative hypotheses, for reduction of hemoglobin, it is left tailed,              
              
Ho:   ud   =   0  
Ha:    ud   <   0   [ANSWER, B]

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33.
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical t,              

As

df = n - 1 =    13          

Then

tcrit =    -1.770933396   [ANSWER]  

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34.
              
Getting the test statistic, as              
              
X = sample mean =    -0.875          
uo = hypothesized mean =    0          
n = sample size =    14          
s = standard deviation =    1.159          
              
Thus, t = (X - uo) * sqrt(n) / s =    -2.824806051          
As t < -1.771, we REJECT HO. [OPTION A]

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35.

              
For the   0.99   confidence level, then      
df =    13          
alpha/2 = (1 - confidence level)/2 =    0.005      
Hence,
  
t(alpha/2) =    3.012275839          
  
Thus,

Margin of error = t(alpha/2) * s / sqrt(n) = 0.93306985 = 0.933

hence,

OPTION B: -0.875 +/- 0.933 [ANSWER]

          

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