A random sample of 50 students at the University of Florida

A random sample of 50 students at the University of Florida revealed the mean number of hours slept last night was 6 hours and 48 minutes (6.8 hours). The standard deviation of the sample was .9 hours. A previous study concluded that Americans get a mean of 7 hours of sleep per night. Is it reasonable to conclude that the students at the University of Florida sleep less than the typical American? Compute the p-value.

THIS NEEDS TO BE IN EXCEL FORMAT

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   7  
Ha:    u   <   7  
              
As we can see, this is a    left   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    6.8          
uo = hypothesized mean =    7          
n = sample size =    50          
s = standard deviation =    0.9          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.571348403          
As this is a left tailed test, we get the left tailed area of this z score using NORMSDIST function. Typing

=NORMSDIST(-1.571348403)
          
We get the P value of
              
p =    0.058050872   [ANSWER]

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At 0.05 level, it may not be reasonable to conclude that the students at the University of Florida sleep less than the typical American. [But at 0.10 level, it will be.]