Test the hypothesis Ho mu 10 against Ha mu 10 based on a ra

Test the hypothesis Ho: mu = 10 against Ha: mu 10 based on a random sample of size n = 16 from a normal population with unknown mean mu and known variance sigma = 5 if the random sample yielded a sample average of 13. Use alpha = .025. Obtain a 35% confidence interval for mu given the sample mean and sample size 7a. Discuss briefly how you\'re your answers to part 7a. and part 7b. are related.

Solution

7a.

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   10  
Ha:    u   =/   10  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical z, as alpha =    0.025   ,      
alpha/2 =    0.0125          
zcrit =    +/-   2.241402728      
              
Getting the test statistic, as              
              
X = sample mean =    13          
uo = hypothesized mean =    10          
n = sample size =    16          
s = standard deviation =    5          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.4          
              
Also, the p value is              
              
p =    0.016395072          
              
As |z| > 2.2414, and P < 0.025, we   REJECT THE NULL HYPOTHESIS.          

Hence, there is significant evidence that the true mean is not equal to 10 at 0.025 level. [CONCLUSION]

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7b.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    13          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    16          
              
Thus,              
Margin of Error E =    2.449954981          
Lower bound =    10.55004502          
Upper bound =    15.44995498          
              
Thus, the confidence interval is              
              
(   10.55004502   ,   15.44995498   ) [ANSWER]

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7C.

We can use an alpha level two tailed test to predict the confidence interval of confidence (1-alpha), and vice versa.

Here, although 0.025 and 0.95 are not complements, we see that they are still consistent, because the interval does not contain 10 as well. Hence, we see that the P value is less than 0.05 as well.