Question

Please use ONLY one Excel file to complete case study one, and use one spreadsheet for each problem. Finally, upload the Excel file to the submission link (Week 3 Case Study) for grading. No credit will be granted for problems that are not completed using Excel. In 2011, when the Gallup organization polled investors, 26% rated gold the best long-term investment. But in April of 2013 Gallup surveyed a random sample of U.S. adults. Respondents were asked to select the best long-term investment from a list of possibilities. Only 189 of the 760 respondents chose gold as the best long-term investment. By contrast, only 87 chose bonds. a. Compute the standard error for each sample proportion. Compute and describe a 90% confidence interval in the context of the question. b. Do you think opinions about the value of gold as a long-term investment have really changed from the old 26% favorability rate, or do you think this is just sample variability? Explain. c. Suppose we want to keep the margin of error at 3%, and we still want to construct a 90% confidence interval. What is the necessary sample size? d. Based on the sample size obtained in part c, suppose 167 respondents chose gold as the best long-term investment. Compute the standard error for choosing gold as the best long-term investment. Compute and describe a 90% confidence interval in the context of the question. e. Based on the results of part d, do you think opinions about the value of gold as a long-term investment have really changed from the old 26% favorability rate, or do you think this is just sample variability? Explain.

Answer 1

a)

for gold rated

Number of Items of Interest,   x =   189
Sample Size,   n =    760
      
Sample Proportion ,    p̂ = x/n =    0.249
  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0157

margin of error , E = Z*SE =    1.645   *   0.0157   =   0.0258
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.249   -   0.0258   =   0.2229
Interval Upper Limit = p̂ + E =   0.249   +   0.0258   =   0.2745
                  
90%   confidence interval is (   0.2229   < p <    0.2745   )

......

for bond proportion

Level of Significance,   α =    0.10  
Number of Items of Interest,   x =   87  
Sample Size,   n =    760  
          
Sample Proportion ,    p̂ = x/n =    0.114  
  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0115  

z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
margin of error , E = Z*SE =    1.645   *   0.0115   =   0.0190
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.114   -   0.0190   =   0.0955
Interval Upper Limit = p̂ + E =   0.114   +   0.0190   =   0.1335
                  
90%   confidence interval is (   0.0955   < p <    0.1335   )

b)

Ho :   p =    0.26                  
H1 :   p ╪   0.26       (Two tail test)          
                          
Level of Significance,   α =    0.10                  
Number of Items of Interest,   x =   189                  
Sample Size,   n =    760                  
                          
Sample Proportion ,    p̂ = x/n =    0.2487                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0159                  
Z Test Statistic = ( p̂-p)/SE = (   0.2487   -   0.26   ) /   0.0159   =   -0.7112
                          
  
p-Value   =   0.47696 [excel formula =2*NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       
There is not enough evidence to conclude that opinions about the value of gold as a long-term investment have really changed from the old 26% favorability rate          

c)

sample proportion ,   p̂ =    0.2486   
sampling error ,    E =   0.03                          
Confidence Level ,   CL=   0.90                          
                                  
alpha =   1-CL =   0.10                          
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.645   /   0.03   ) ² *   0.25   * ( 1 -   0.25   ) =    561.7
                                  
                                  
so,Sample Size required=       562                  

d)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   167          
Sample Size,   n =    562          
                  
Sample Proportion ,    p̂ = x/n =    0.297          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0193          

margin of error , E = Z*SE =    1.645   *   0.0193   =   0.0317
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.297   -   0.0317   =   0.2654
Interval Upper Limit = p̂ + E =   0.297   +   0.0317   =   0.3289
                  
90%   confidence interval is (   0.2654   < p <    0.3289   )

e)

0.26 does not lie in the confidence interval so opinions about the value of gold as a long-term investment have really changed from the old 26% favorability rate.