Consider the following hypothesis statement using 010 and d
Consider the following hypothesis statement using = 0.10 and data from two independent samples. Assume the population variances are equal and the populations are normally distributed. Complete parts a and b.
H0: 1 2 11
H1: 1 2 > 11
overbarX1 = 66.6 overbarx2 = 52.1
s1 = 16.5 s2 = 17.1
n1 = 19 n2 = 22
a) Calculate the appropriate test statistic and interpret the result.
The test statistic is ______ (round to 2 decimal places)
The critical value(s) is (are) ________ (round to 2 decimal places)
Because the test statistic falls / does not fall / is less than / is greater than , reject / do not reject the null hypothesis.
b) The p-value is _____ (round to 3 decimal places)
a)
A.
Since the p-value is
lessless
than the significance level,
rejectreject
the null hypothesis.
B.
Since the p-value is less / not less than the significance level, reject / do not reject the null hypothesis.
| H0: 1 2 11 | 
Solution
a)
Formulating the null and alternative hypotheses,              
               
 Ho:   u1 - u2   <=   11  
 Ha:   u1 - u2   >   11  
 At level of significance =    0.1          
 As we can see, this is a    right   tailed test.      
 Calculating the means of each group,              
               
 X1 =    66.6          
 X2 =    52.1          
               
 Calculating the standard deviations of each group,              
               
 s1 =    16.5          
 s2 =    17.1          
               
 Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
               
 n1 = sample size of group 1 =    19          
 n2 = sample size of group 2 =    22          
Also, sD =    5.255502926          
               
 Thus, the t statistic will be              
               
 z = [X1 - X2 - uD]/sD =    0.665968614 = 0.67   [ANSWER, TEST STATISTIC]
******************************************      
               
 where uD = hypothesized difference =    11          
               
 Now, the critical value for Z is              
               
 zcrit =    1.28 [ANSWER]
****************************************  
               
 Because the test statistic is LESS THAN the critical value , DO NOT REJECT the null hypothesis.
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b)
The P value for the right tailed area of z = 0.665968614 is
P = 0.252715581. [ANSWER]
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B. Since the p-value is NOT LESS than the significance level, DO NOT reject the null hypothesis.


