H0 394 HA 394 A sample of 31 observations yields a sample m

H0: 39.4 HA: > 39.4

A sample of 31 observations yields a sample mean of 40.4. Assume that the sample is drawn from a normal population with a known population standard deviation of 5.9. Use Table 1.

a. Calculate the p-value. (Round \"z\" value to 2 decimal places.)

b. What is the conclusion if = 0.01?

c. Calculate the p-value if the above sample mean was based on a sample of 140 observations. (Round \"z\" value to 2 decimal places.)

d. What is the conclusion if = 0.01?

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   39.4  
Ha:    u   >   39.4  
              
As we can see, this is a    right   tailed test.      
              
Getting the test statistic, as              
              
X = sample mean =    40.4          
uo = hypothesized mean =    39.4          
n = sample size =    31          
s = standard deviation =    5.9          
              
Thus, z = (X - uo) * sqrt(n) / s =    0.943688875 = 0.94          
              
Also, the p value is              
              
p =    0.1736 [ANSWER]

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b)

As P > 0.01, we FAIL TO REJECT HO.

There is no significant evidence that the true mean is greater than 39.4. [ANSWER]

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c)

Getting the test statistic, as              
              
X = sample mean =    40.4          
uo = hypothesized mean =    39.4          
n = sample size =    140          
s = standard deviation =    5.9          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.005450774 = 2.01          
              
Also, the p value is              
              
p =    0.022215594 [ANSWER]

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d)

STILL, as P > 0.01, WE FAIL TO REJECT HO.
          
              
There is no significant evidence that the true mean is greater than 39.4. [ANSWER]
  

          
              
Comparing z and zcrit (or, p and significance level), we   FAIL TO REJECT THE NULL HYPOTHESIS.