Applet Exercise How was Figure 9.1 obtained? Access the

Chapter 9, Problem 9E

(choose chapter or problem)

Applet Exercise How was Figure  obtained? Access the applet PointSingle at www. thomsonedu.com/statistics/wackerly. The top applet will generate a sequence of Bernoulli trials \(\left[X_{i}=1,0 \text { with } p(1)=p, p(0)=1-p\right]\) with \(p=.5\), a scenario equivalent to successively tossing a balanced coin. Let \(Y_{n}=\sum_{i=1}^{n} X_{i}=\) the number of \(1 s\) in the first  trials and \(\hat{p}_{n}=Y_{n} / n\). For each , the applet computes \(\hat{p}_{n}\) and plots it versus the value of .
                                                                                                        a If \(\hat{p}_{5}=2 / 5\), what value of \(X_{6}\) will result in \(\hat{p}_{6}>\hat{p}_{5}\) ?
b Click the button "One Trial" a single time. Your first observation is either 0 or 1 . Which value did you obtain? What was the value of \(\hat{p}_{1}\) ? Click the button "One Trial" several more times. How many trials
n have you simulated? What value of ˆpn did you observe? Is the value close to .5, the true value of p? Is the graph a flat horizontal line? Why or why not?                                      c Click the button “100 Trials” a single time. What do you observe? Click the button “100 Trials” repeatedly until the total number of trials is 1000. Is the graph that you obtained identical to the one given in Figure 9.1? In what sense is it similar to the graph in Figure 9.1?                        d Based on the sample of size 1000, what is the value of \(\hat{p} 1000\)? Is this value what you expected to observe?                                                                                        e Click the button “Reset.” Click the button “100 Trials” ten times to generate another sequence of values for \(\hat{p}\). Comment.

Equation Transcription:

 

  

 

 

 

   

     

 

     

   

Text Transcription:

\left[X_i=1,0 with  p(1)=p, p(0)=1-p\right]

p=.5

Y_n=\sum_i=1^n X_i =

1 s  

\hatp_n=Y_n / n

\hat{p}_{n}

\hat{p}_{5}=2 / 5

X6  

\hat p_ 6 >\hat p_5

\hat p_1

\hat p 1000

\hat p

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