Question

41.

A poll showed that

5757​%

of randomly surveyed adults in a certain country said their country benefits from having a rich class. Complete parts​ (a) through​ (h) below.a. Assuming the sample size was

400400​,

how many would have said that their country benefits from having a rich​ class?Assuming the sample size was

400400​,

nothing

people in the sample would have said that their country benefits from having a rich class.

​(Round to the nearest integer as​ needed.)

b. Is the sample size large enough to apply the Central Limit​ Theorem? Explain. Assume the other conditions for using the CLT are met.

Select the correct choice below and fill in the answer​ box(es) to complete your choice.

​(Round to the nearest integer as​ needed.)

A.​Yes, the sample size is large​ enough, since the expected number of successes is

nothing

and the expected number of failures is

nothing​,

both of which are greater than or equal to 10.

B.​Yes, the sample size is large​ enough, since the estimated standard error is

nothing​,

which is greater than or equal to 10.

C.​No, the sample size is not large​ enough, since the expected number of successes is

nothing

and the expected number of failures is

nothing​,

at least one of which is less than 10.

D.​No, the sample size is not large​ enough, since the estimated standard error is

nothing​,

which is less than 10.

c. Find a​ 95% confidence interval for the percent that believe that their country benefits from having a rich​ class, using the numbers from part​ (a).

The​ 95% confidence interval is

left parenthesis nothing % comma nothing % right parenthesis%,%.

​(Round to one decimal place as​ needed.)

d. Find the width of the interval you found in part​ (c) by subtracting the lower boundary from the upper boundary.

The width of the interval is

nothing​%.

​(Round to one decimal place as​ needed.)

e. Now assuming the sample size was multiplied by

99

​(nequals=36003600​)

and the percentage was still

5757​%,

how many would have said their country benefits from having a rich​ class?Assuming the sample size was

36003600​,

nothing

would have said they were thriving.

​(Round to the nearest integer as​ needed.)

f. Find a​ 95% confidence​ interval, using the numbers from part​ (e).

The​ 95% confidence interval is

left parenthesis nothing % comma nothing % right parenthesis%,%.

​(Round to one decimal place as​ needed.)

g. What is the width of the interval you found in part​ (f)?

The width of the interval is

nothing​%.

​(Round to one decimal place as​ needed.)

h. When the sample size is multiplied by

99

is the width of the interval divided by

99​?

If​ not, what is it divided​ by? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.​No, when the sample size is multiplied by

99​,

the width of the interval is divided by

nothing.

​(Round to the nearest integer as​ needed.)

B.​Yes, when the sample size is multiplied by

99​,

the width of the interval is divided by

99.

Click to select your answer(s).

Answer 1

a) np = 400*0.57 = 228

b)

A.​Yes, the sample size is large​ enough, since the expected number of successes is 228

and the expected number of failures is (400-228) = 172

both of which are greater than or equal to 10.

c)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   228          
Sample Size,   n =    400          
                  
Sample Proportion ,    p̂ = x/n =    0.5700          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0248          
margin of error , E = Z*SE =    1.960   *   0.0248   =   0.0485
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.570   -   0.0485   =   0.5215
Interval Upper Limit = p̂ + E =   0.570   +   0.0485   =   0.6185
                  
95%   confidence interval is ( 52.1% < p < 61.9% )

d)The width of the interval is 9.7%

e) np=3600*0.57=2052

f)

Level of Significance,   α =    0.05          
Sample Size,   n =    3600          
                  
Sample Proportion ,    p̂ = x/n =    0.5700          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0083          
margin of error , E = Z*SE =    1.960   *   0.0083   =   0.0162
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.570   -   0.0162   =   0.554
Interval Upper Limit = p̂ + E =   0.570   +   0.0162   =   0.586
                  
95%   confidence interval is ( 55.4% < p < 58.6% )

g) width = 3.2%

h) A.​No, when the sample size is multiplied by 9 ,the width of the interval is divided by 3