Question

you are to create one 90%, one 95%, and one 99.7% confidence interval for the true proportion of online students

Raw data:

Survey Question: Do you have kids under the age of 18?

Yes - 15

No - 44

Total - 59

1. Summary of Data

2. A statement of what your variable X means

3. A statement of what your statistic means (with proper notation).

4. Your calculated statistic

5. The distribution for your variable (with proper notation)

6. Your check for normality

7. A statement of the what your population parameter means

8. Your work (Yup, show the steps out)

9. Your confidence interval

10. A conclusion

Answer 1

1)

Sample Size,   n =    59
Number of Items of Interest,   x =   44

Sample Proportion ,    p̂ = x/n =    0.7458

................

x = number of students who are online (who are above 18 will be online only)

.........

z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0567         

................  

normality check

np>5 so , 59*0.7458 =44.002 > 5

n(1-p)>5 =59*(1-0.7458) =14.9978 > 5

.................

population parameter means range within which the proportion of online students lie

..............

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   44          
Sample Size,   n =    59          
                  
Sample Proportion ,    p̂ = x/n =    0.746          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0567          
margin of error , E = Z*SE =    1.645   *   0.0567   =   0.0932
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.746   -   0.0932   =   0.6525
Interval Upper Limit = p̂ + E =   0.746   +   0.0932   =   0.8390
                  
90%   confidence interval is (   0.6525   < p <    0.8390   )

..........

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   44          
Sample Size,   n =    59          
                  
Sample Proportion ,    p̂ = x/n =    0.746          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0567          
margin of error , E = Z*SE =    1.960   *   0.0567   =   0.1111
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.746   -   0.1111   =   0.6347
Interval Upper Limit = p̂ + E =   0.746   +   0.1111   =   0.8569
                  
95%   confidence interval is (   0.6347   < p <    0.8569   )

.............

Level of Significance,   α =    0.003          
Number of Items of Interest,   x =   44          
Sample Size,   n =    59          
                  
Sample Proportion ,    p̂ = x/n =    0.746          
z -value =   Zα/2 =    2.968   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0567          
margin of error , E = Z*SE =    2.968   *   0.0567   =   0.1682
                  
99.7%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.746   -   0.1682   =   0.5775
Interval Upper Limit = p̂ + E =   0.746   +   0.1682   =   0.9140
                  
99.7%   confidence interval is (   0.5775   < p <    0.9140   )

.....

lower limit of interval of true proportion decreases when we increase confidence level

and UPPER limit of interval decreases as we increase the confidence level


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