Question

A survey was undertaken by Bruskin/Goldring Research for Quicken to determine how people plan to meet their financial goals in the next year. Respondents were allowed to select more than one way to meet their goals. Thirty-one percent said that they were using a financial planner to help them meet their goals. Twenty-four percent were using family/friends to help them meet their financial goals followed by broker/accountant (19%), computer software (17%), and books (14%). Suppose another researcher takes a similar survey of 550 people to test these results.

a)  If 183 people respond that they are going to use a financial planner to help them meet their goals, is this proportion enough evidence to reject the 31% figure generated in the Bruskin/Goldring survey using α = 0.10.

The value of the test statistic is ________ & reject the null hypothesis/ fait to reject the null hypothesis

There is/ is not enough evidence to declare that the proportion is any different/ not any different than 0.31

b)  If 143 respond that they are going to use family/friends to help them meet their financial goals, is this result enough evidence to declare that the proportion is significantly lower than Bruskin/Goldring’s figure of 0.24 if α = 0.05?

The value of the test statistic is ________ & reject the null hypothesis/ fait to reject the null hypothesis

There is/ is not enough evidence to declare that the proportion is any different/ not any different than 0.24

Answer 1

a)

Ho :   p =    0.31                  
H1 :   p ╪   0.31       (Two tail test)          
                          
Level of Significance,   α =    0.10                  
Number of Items of Interest,   x =   183                  
Sample Size,   n =    550                  
                          
Sample Proportion ,    p̂ = x/n =    0.3327                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0197                  
Z Test Statistic = ( p̂-p)/SE = (   0.3327   -   0.31   ) /   0.0197   =   1.1525
                          
  
p-Value   =   0.2491   [excel formula =2*NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       

The value of the test statistic is ____1.1525____ & fail to reject the null hypothesis

There is not enough evidence to declare that the proportion is any different than 0.31

b)

Ho :   p =    0.24                  
H1 :   p <   0.24       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   143                  
Sample Size,   n =    550                  
                          
Sample Proportion ,    p̂ = x/n =    0.2600                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0182                  
Z Test Statistic = ( p̂-p)/SE = (   0.2600   -   0.24   ) /   0.0182   =   1.0982
                          
  
p-Value   =   0.8640   [excel function =NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       

The value of the test statistic is ____1.0982____ & fait to reject the null hypothesis

There is not enough evidence to declare that the proportion is any different  than 0.24