A recent national survey found that high school students watched an average (mea

Question

A recent national survey found that high school students watched an average (mean) of 6.8 DVDs per month with a population standard deviation of 1.8 DVDs. The distribution of DVDs watched per month follows the normal distribution. A random sample of 36 college students revealed that the mean number of DVDs watched last month was 6.2. At the .05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round your answer to 1 decimal place.)

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round your answer to 1 decimal place.)

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   6.8  
Ha:    u   <   6.8  
              
As we can see, this is a    left   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    6.2          
uo = hypothesized mean =    6.8          
n = sample size =    36          
s = standard deviation =    1.8          
              
Thus, z = (X - uo) * sqrt(n) / s =    -2.0   [ANSWER, Z STATISTIC]      
              

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Other info:

Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      

Also, the p value is              
              
p =    0.022750132          
              
Comparing z and zcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.          
              

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