Use the oneproportion ztest and zinterval procedure to condu

Use the one-proportion z-test and z-interval procedure to conduct the required hypothesis and specified confidence interval.

Solution

2.

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.2          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.063245553          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.123959006          
lower bound = p^ - z(alpha/2) * sp =   0.076040994          
upper bound = p^ + z(alpha/2) * sp =    0.323959006          
              
Thus, the confidence interval is              
              
(   0.076040994   ,   0.323959006   ) [ANSWER]

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b)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   <=   0.65
Ha:   p   >   0.65
As we see, the hypothesized po =   0.65      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.7      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.067453688      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    0.741249317      
          
As this is a    1   tailed test, then, getting the p value,  
          
p =    0.229271143      
significance level =    0.1      

As P > 0.1, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

There is no significant evidence that p > 0.65 at 0.10 level. [CONCLUSION]