Solution Found!
Of n1 randomly selected engineering students at ASU,X1
Chapter 7, Problem 39E(choose chapter or problem)
Of \(n_{1}\) randomly selected engineering students at ASU, \(X_{1}\) owned an HP calculator, and of \(n_{2}\) randomly selected engineering students at Virginia Tech, \(X_{2}\) owned an HP calculator. Let \(p_{1}\) and \(p_{2}\) be the probability that randomly selected ASU and Virginia Tech engineering students, respectively, own HP calculators.
(a) Show that an unbiased estimate for \(p_{1}-p_{2}\) is \(\left(X_{1} / n_{1}\right)=X_{2} / n_{2}\).
(b) What is the standard error of the point estimate in part (a)?
(c) How would you compute an estimate of the standard error found in part (b)?
(d) Suppose that \(n_1=200,\ X_1=150,\ n_2=250,\) and \(X_{2}=185\). Use the results of part (a) to compute an estimate of \(p_{1}-p_{2}\).
(e) Use the results in parts (b) through (d) to compute an estimate of the standard error of the estimate.
Equation Transcription:
Text Transcription:
n_1
X_1
n_2
X_2
p_1
p_2
p_1-p_2
(X_1/n_1)=X_2/n_2
n_1=200, X_1=150, n_2=250,
X_2=185
p_1-p_2
Questions & Answers
QUESTION:
Of \(n_{1}\) randomly selected engineering students at ASU, \(X_{1}\) owned an HP calculator, and of \(n_{2}\) randomly selected engineering students at Virginia Tech, \(X_{2}\) owned an HP calculator. Let \(p_{1}\) and \(p_{2}\) be the probability that randomly selected ASU and Virginia Tech engineering students, respectively, own HP calculators.
(a) Show that an unbiased estimate for \(p_{1}-p_{2}\) is \(\left(X_{1} / n_{1}\right)=X_{2} / n_{2}\).
(b) What is the standard error of the point estimate in part (a)?
(c) How would you compute an estimate of the standard error found in part (b)?
(d) Suppose that \(n_1=200,\ X_1=150,\ n_2=250,\) and \(X_{2}=185\). Use the results of part (a) to compute an estimate of \(p_{1}-p_{2}\).
(e) Use the results in parts (b) through (d) to compute an estimate of the standard error of the estimate.
Equation Transcription:
Text Transcription:
n_1
X_1
n_2
X_2
p_1
p_2
p_1-p_2
(X_1/n_1)=X_2/n_2
n_1=200, X_1=150, n_2=250,
X_2=185
p_1-p_2
ANSWER:
Step 1 of 7
(a)
The expected value (or mean) of a binomial distribution is the product of the sample size and the probability of success:
Step 3 of 7
Thus 
