Randomly selected 60 student cars have ages with a mean of 7.8 years and a stand

Question

Randomly selected 60 student cars have ages with a mean of 7.8 years and a standard deviation of 3.4 years, while randomly selected 105 faculty cars have ages with a mean of 6 years and a standard deviation of 3.3 years. 1. Use a 0.05 significance level to test the claim that student cars are older than faculty cars. The test statistic is The critical value is Is there sufficient evidence to support the claim that student cars are older than faculty cars? A. Yes B. No 2. Construct a 95% confidence interval estimate of the difference 12, where 1 is the mean age of student cars and 2 is the mean age of faculty cars.

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  
At level of significance =    0.05          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    7.8          
X2 =    6          
              
Calculating the standard deviations of each group,              
              
s1 =    3.4          
s2 =    3.3          
              
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 =    105          
Thus, df = n1 + n2 - 2 =    163          
Also, sD =    0.54440881          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    3.306338852          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =        1.654255585      
              
Also, using p values,              
              
p =    0.000581358          
              
As t > 1.654, and P < 005,    WE REJECT THE NULL HYPOTHESIS.          

Hence, YES, there is sufficient evidence to support the claim that student cars are older than faculty cars. [ANSWER, YES]

*********************************************

b)

For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
t(alpha/2) =    2.606327515          
  
Hence,
          
lower bound = [X1 - X2] - t(alpha/2) * sD =    0.381092339          
upper bound = [X1 - X2] + t(alpha/2) * sD =    3.218907661          
              
Thus, the confidence interval is              
              
(   0.381092339   ,   3.218907661   ) [ANSWER]

*****************************************************

Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.