According to the results of the 2004 presidential election, 59% of New Yorkers v

Question

According to the results of the 2004 presidential election, 59% of New Yorkers voted for Kerry. Imagine the set of all possible samples of size 100 from all New Yorkers who voted. For each sample of size 100, the value of the sample proportion P Unit vector is the proportion of the people in the sample that voted for Kerry. Use 3 decimal places. What would the mean of all these sample proportions be? What would the standard deviation of all these sample proportions be? Would the distribution of all these sample proportions be Normal? No Yes Now suppose you took a SRS of size 100 from all New Yorkers who voted. What is the probability that more than 65% of the people in your sample would have voted for Senator Kerry? What is the probability that between 55% and 65% of the people in your sample would have voted for Senator Kerry? The middle 90% of all sample proportions p Unit vector fall between and.

Explanation / Answer

a)

u = mean = p =    0.59 [ANSWER]

*****************
b)
  
s = standard deviation = sqrt(p(1-p)/n) =    0.049183331 [ANSWER]

**********************

c)

Yes, as np(1-p) = 24.19 > 5.

*******************

d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.65      
u = mean = p =    0.59      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.049183331      
          
Thus,          
          
z = (x - u) / s =    1.219925519      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.219925519   ) =    0.111246555 [ANSWER]

********************

e)

Here,          
n =    100      
p =    0.59      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.55      
x2 = upper bound =    0.65      
u = mean = p =    0.59      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.049183331      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.813283679      
z2 = upper z score = (x2 - u) / s =    1.219925519      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.208027716      
P(z < z2) =    0.888753445      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.680725728   [ANSWER]

*******************

f)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.9      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.05      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.644853627      
By symmetry,          
z2 =    1.644853627      
          
As          
          
u = mean =    0.59      
s = standard deviation =    0.049183331      
          
Then          
          
x1 = u + z1*s =    0.50910062   [ANSWER]  
x2 = u + z2*s =    0.67089938   [ANSWER]  
  

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.