Question

The Arizona Department of Transportation (ADOT) is considering changing statewide speed limits from 65 mph to 75 mph. They wish to perform analysis on p, the proportion of Arizona residents that favor this change. In order to prove that a majority of residents favor the change, ADoT wishes to test the following hypotheses using a level of significance of 0.05. Ho:p = 0.5 H1: p > 0.5 ADOT wishes to design the test to have a power of 90% if the true population proportion is equal to 0.55. Part a (5 points): How many residents should be surveyed? You may use this sample size for the remainder of this problem. (Note: You should get a value between 800 and 900 on this part. If you did not, then choose a value between 800 and 900 and use it to work the remainder of the problem.) Part b (4 points): Suppose the sample size found in part a is implemented and 461 of the surveyed residents favor the change. Construct a point estimate on the proportion of interest. Part c (6 points): Identify the test statistic and critical region for the test. State the conclusion for this test. Part d (5 points): What is the P-value for this test? State the conclusion for this test. Part e (5 points): Construct an appropriate confidence interval to verify your conclusion from part c. Part f (5 points): We wish to use a 95% lower confidence interval to estimate p. How many residents should be surveyed so that the error in our estimate is less than 4%.

Answer 1

a)

true proportion ,p=   0.55  
hypothesized proportion ,po =    0.5  
α=   0.050  
power= 1- ß =   0.9  
ß =    0.1  
      
Zα=   1.645   [excel function: =normsinv(α) ]
Z (ß ) =    1.282   [excel function: =normsinv(ß)
      
n = po*(1-po)[Z(α) + Z(ß) ]² / (p-po)² =   0.5*0.5(1.282+1.645)²/(0.55-0.5)² =    856.38
so, sample size=   857  

b) point estimate= Sample Proportion ,    p̂ = x/n =    0.5379

c) Standard Error ,    SE = √( p(1-p)/n ) =    0.0171  
Z Test Statistic = ( p̂-p)/SE =    (0.5379-0.5)/0.0171=   2.220  
          
critical z value =        1.645   [Excel function =NORMSINV(α)

reject Ho if Z > 1.645

decision: reejct ho

d)

p-Value   =   0.0132
Decision:   p-value<α , reject null hypothesis   

e)

Number of Items of Interest,   x =   461      
Sample Size,   n =    857      
Sample Proportion ,    p̂ = x/n =    0.5379      
              
Level of Significance,   α =    0.05      
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0170      
margin of error , E = Z*SE =    1.96*0.017=   0.0334      
              
Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.5379-0.0334=   0.5045      
Interval Upper Limit = p̂ + E =   0.5379+0.0334=   0.5713      
              
95% Confidence interval is (   0.505   < p <    0.571   )
since, CI does not contain 0.50, reject Ho