In order to conduct a hypothesis test of the population mean
In order to conduct a hypothesis test of the population mean, a random sample of 9 observations is drawn from a normally distributed population. The resulting mean and the standard deviation are calculated as 16.9 and 1.9, respectively.
Use the critical value approach to conduct the following tests at = 0.01.
H0: 15.3 against HA: > 15.3
a-1.
Calculate the value of the test statistic. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
a-2.
Calculate the critical value. (Round your answer to 3 decimal places.)
a-3.
What is the conclusion?
H0: = 15.3 against HA: 15.3
b-1.
Calculate the value of the test statistic. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
b-2.
Calculate the critical value(s). (Round your answers to 3 decimal places.)
| H0: 15.3 against HA: > 15.3 | 
Solution
a-1)
Formulating the null and alternative hypotheses,              
               
 Ho:   u   <=   15.3  
 Ha:    u   >   15.3  
               
 As we can see, this is a    right   tailed test.      
               
               
 Getting the test statistic, as              
               
 X = sample mean =    16.9          
 uo = hypothesized mean =    15.3          
 n = sample size =    9          
 s = standard deviation =    1.9          
               
 Thus, t = (X - uo) * sqrt(n) / s =    2.526315789 = 2.53 [ANSWER]
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A-2)
   
 Thus, getting the critical t,              
 df = n - 1 =    8          
 tcrit = 2.896 [ANSWER]
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a-3)
OPTION D: Do not reject H0 since the value of the test statistics is smaller than the critical value.
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b-1)
Formulating the null and alternative hypotheses,              
               
 Ho:   u   =   15.3  
 Ha:    u   =/   15.3  
               
 As we can see, this is a    two   tailed test.      
               
 Getting the test statistic, as              
               
 X = sample mean =    16.9          
 uo = hypothesized mean =    15.3          
 n = sample size =    9          
 s = standard deviation =    1.9          
               
 Thus, t = (X - uo) * sqrt(n) / s =    2.526315789 = 2.53 [ANSWER]
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b-2)
Thus, getting the critical t,              
 df = n - 1 =    8          
 tcrit =    +/-   3.355 [ANSWER]      
               


