*PLEASE HELP I\'M STUCK* A school principal was interested in whether students i

Question

*PLEASE HELP I'M STUCK*

A school principal was interested in whether students in the chess club scored above or below the mean grade point average of her school. She figured out the average GPA at the school to be 2.55 with a standard deviation of .5 and noted that the distribution was approximately normal. She then figured out the average GPA of 15 students in the chess club. Their mean GPA was 2.76. Did this group represent a population different from the students in general at this school? (use the .05 significance level)

A) What type of statistical test should be conducted on this data and why? (Z test, T test for single sample, T test for dependent means, T test for independent means, or Correlation)

B) Conducd the 5 steps of hypothesis testing and show your work

C) Figure the confidence limits for the 95% confidence interval (Show work)

Explanation / Answer

a)

We only have 1 sample, the chess club, and it is a small sample (n = 15), so we use T TEST. [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   2.55  
Ha:    u   =/   2.55  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    14          
tcrit =    +/-   2.144786688      
              
Getting the test statistic, as              
              
X = sample mean =    2.76          
uo = hypothesized mean =    2.55          
n = sample size =    15          
s = standard deviation =    0.5          
              
Thus, t = (X - uo) * sqrt(n) / s =    1.626653005          
              
Also, the p value is              
              
p =    0.126100658          
              
As |t| < 2.145, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

Hence, there is no significant difference from the mean GPA of the chess club and the students in general at this school, at 0.05 level. [CONCLUSION]

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c)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    2.76          
t(alpha/2) = critical t for the confidence interval =    2.144786688          
s = sample standard deviation =    0.5          
n = sample size =    15          
df = n - 1 =    14          
Thus,              
Margin of Error E =    0.276890771          
Lower bound =    2.483109229          
Upper bound =    3.036890771          
              
Thus, the confidence interval is              
              
(   2.483109229   ,   3.036890771   ) [ANSWER]

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