Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbarx​, is found to be 115​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 95​% confidence interval about muμ if the sample​ size, n, is 21. ​(b) Construct a 95​% confidence interval about muμ if the sample​ size, n, is 26. ​(c) Construct a 96​% confidence interval about muμ if the sample​ size, n, is 21. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed? . ​(a) Construct aa 95​% confidence interval about muμ if the sample​ size, n, is 21. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) ​(b) Construct aa 95​% confidence interval about muμ if the sample​ size, n, is 26. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) How does increasing the sample size affect the margin of​ error, E?

A. As the sample size increases, the margin of error increases.

B. As the sample size increasesincreases​, the margin of error stays the same.

C. As the sample size increases, the margin of error decreases. ​(c) Construct a 96​% confidence interval about muμ if the sample​ size, n, is 21. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E? A. As the level of confidence increases​, the size of the interval decreases. B. As the level of confidence increases​, the size of the interval increases.

Answer 1

a)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   20          
't value='   tα/2=   2.0860   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.0000   / √   21   =   2.182179
margin of error , E=t*SE =   2.0860   *   2.18218   =   4.551945
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    115.00   -   4.551945   =   110.448055
Interval Upper Limit = x̅ + E =    115.00   -   4.551945   =   119.551945
95%   confidence interval is (   110.45   < µ <   119.55   )

b)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   25          
't value='   tα/2=   2.0595   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.0000   / √   26   =   1.961161
margin of error , E=t*SE =   2.0595   *   1.96116   =   4.039087
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    115.00   -   4.039087   =   110.960913
Interval Upper Limit = x̅ + E =    115.00   -   4.039087   =   119.039087
95%   confidence interval is (   110.96   < µ <   119.04   )

c)

Level of Significance ,    α =    0.04          
degree of freedom=   DF=n-1=   20          
't value='   tα/2=   2.1967   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.0000   / √   21   =   2.182179
margin of error , E=t*SE =   2.1967   *   2.18218   =   4.793500
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    115.00   -   4.793500   =   110.206500
Interval Upper Limit = x̅ + E =    115.00   -   4.793500   =   119.793500
96%   confidence interval is (   110.21   < µ <   119.79   )

d)

No we could not have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed

C. As the sample size increases, the margin of error decreases.

As the level of confidence increases​, the size of the interval increases.

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