Question

In 1995, a random sample of 100 adults that had investments in the stock market found that only 20 said they were investing for the long haul rather than to make quick profits. A simple random sample of 100 adults that had investments in the stock market in 2002 found that 36 were investing for the long haul rather than to make quick profits. Let p1995 and p2002 be the actual proportion of all adults with investments in the stock market in 1995 and in 2002, respectively, that were investing for the long haul.

a. What are the 1995 and 2002 estimated proportions of all adults with long haul investments in the stock market?

b. State the null and alternative hypotheses if the researcher is interested to see if nothing changed when it comes to the proportions of adults investing for the long haul between 1995 and 2002. 6

c. Conduct the hypothesis test from point b. at α=0.05 and formulate the conclusion using the observed vs critical value method.

d. State the null and alternative hypotheses for the case in which the researcher wants to test if the proportion of adults investing for the long haul increase over time. e. Conduct the hypothesis test from point d. at α=0.05 using the p-value method.

f. Construct the 95% confidence interval for the difference in proportions associated with the hypothesis test from b.

g. What is the margin of error for a 97% confidence interval for the difference in proportions associated with the hypothesis test from b? How does.it compare with the margin of error from point f. Explain the differences.

Answer 1

a)

proportion success of sample 1 , p̂1=   x1/n1=   0.2000
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.360

b)

Ho:   p1 - p2 =   0
Ha:   p1 - p2 ╪   0

c)

difference in sample proportions, p̂1 - p̂2 =     0.2000   -   0.3600   =   -0.1600
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2800          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0635          
Z-statistic = (p̂1 - p̂2)/SE = (   -0.160   /   0.0635   ) =   -2.5198
                  
z-critical value , Z* =        1.9600   [excel formula =NORMSINV(α/2)]      

decision :    |test stat| > |critical value|,Reject null hypothesis

......

d)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 <   0          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2000   -   0.3600   =   -0.1600
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2800          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0635          
Z-statistic = (p̂1 - p̂2)/SE = (   -0.160   /   0.0635   ) =   -2.5198
                  
p-value =        0.0059   [Excel function =NORMSDIST(z)      
decision :    p-value<α,Reject null hypothesis

conclusion :the proportion of adults investing for the long haul increase over time

..........

f)

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0625          
margin of error , E = Z*SE =    1.960   *   0.0625   =   0.1225
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.160   -   0.1225   =   -0.2825
upper limit = (p̂1 - p̂2) + E =    -0.160   +   0.1225   =   -0.0375
                  
so, confidence interval is (   -0.2825   < p1 - p2 <   -0.0375   )  

.........

g)

level of significance, α =   0.03              
Z critical value =   Z α/2 =    2.170   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0625        
margin of error , E = Z*SE =    2.170   *   0.0625   =   0.1356

thanks

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