Many Students have been conducted to test the effects of Mar

Many Students have been conducted to test the effects of Marijuana use on mental abilities. In one such study, group of light and heavy users of marijuana in college were tested for memory recall, with the results given as follows: light users = n= 70, x1 = 49.6, s1 = 8.2 heavy users = n=60, x2 = 46.9, s2 = 5.9. Use a 0.02 sitgnificance level to test the claim that the population of heavy users has a lower mean than light users. Should marijuana use be of concern to college student? Construct a confidence interval for the difference between the two population. Does the confidence interval include 0? What does it suggest about the equality of the two population? Interpret the interval.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0  

At level of significance =    0.02          

As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    46.9          
X2 =    49.6          
              
Calculating the standard deviations of each group,              
              
s1 =    5.9          
s2 =    8.2          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    60          
n2 = sample size of group 2 =    70          
Thus, df = n1 + n2 - 2 =    128          
Also, sD =    1.241264716          
              
Thus, the z statistic will be              
              
z = [X1 - X2 - uD]/sD =    -2.175200797          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for z is              
              
zcrit = -2.053748911

              
As |z| > 2.0537,    WE REJECT THE NULL HYPOTHESIS.          
              
Also, using p values,              
              
p =    0.015726303          
              
As P < 0.02,    WE REJECT THE NULL HYPOTHESIS.

Thus, there is significant evidence that the mean of heavy users is less than the mean of light users. [CONCLUSION]

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For the   0.98   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.01          
Z(alpha/2) =    2.053748911          
                  
upper bound = [X1 - X2] + t(alpha/2) * sD = -0.150753942

Thus,

u(light) - u(heavy) < -0.150753942. [ANSWER]

This interval does not include 0. [ANSWER]

This suggests that the mean difference is totally less than 0, that there is significant difference in the mean scores.

We are 98% confident that light users have a mean at least greater by 0.150753942.