A local college requires an English composition course for a

A local college requires an English composition course for all freshmen. This year they are evaluating a new online version of the course. A random sample of n = 16 freshmen is selected, and the students are placed in the online course. At the end of the semester, all freshmen take the same English composition exam. The average score for the sample is M = 76. For the general population of freshmen who took the traditional lecture class, the exam scores form a normal distribution with a mean of = 80. If the Final exam scores for the population have a standard deviation of sigma = 12, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Assume a two-tailed test with alpha = .05. These results indicate: Failure to reject the null hypothesis; the course has a significant effect Rejection of the null hypothesis; the course does not have a significant effect Rejection of the null hypothesis; the course has a significant effect Failure to reject the null hypothesis; the course does not have a significant effect If the population standard deviation is sigma = 6, is the sample sufficient to demonstrate a significant difference? Again, assume a two-tailed test with alpha = .05. These results indicate: Rejection of the null hypothesis; the course does not have a significant effect Failure to reject the null hypothesis; course has a significant effect Rejection of the null hypothesis; course has a significant effect Failure to reject the null hypothesis; course does not have a significant effect. Comparing your answers for the above parts, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test. A larger standard deviation produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis. A smaller standard deviation produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   80  
Ha:    u   =/   80  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha/2 =    0.025          
zcrit =    +/-   1.959963985 [ANSWER, Z CRITICAL]

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Getting the test statistic, as              
              
X = sample mean =    76          
uo = hypothesized mean =    80          
n = sample size =    16          
s = standard deviation =    12          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.333333333   [answer, Z]

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As |z| < 1.95996, we choose

OPTION D: FAILURE TO REJECT THE NULL HYPOTHESIS; The course does not have a significant effect.

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,      
alpha/2 =    0.025          
zcrit =    +/-   1.959963985   [ANSWER, Z CRITICAL]

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Getting the test statistic, as              
              
X = sample mean =    76          
uo = hypothesized mean =    80          
n = sample size =    16          
s = standard deviation =    6          
              
Thus, z = (X - uo) * sqrt(n) / s =    -2.666666667   [ANSWER, Z]

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As |z| > 1.95996, we choose

OPTION c: REJECTION OF THE NULL HYPOTHESIS; The course have a significant effect.

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OPTION A: A larger standard deviation produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis.