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To deal with issues such as the distribution of p not

Chapter 5, Problem 43AYU

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QUESTION:

Problem 43AYU

To deal with issues such as the distribution of p not following a normal distribution A. Agresti and B. Coull (Approximate Is Better Than “Exact” for Interval Estimation of Binomial Proportion. American Statistician, 52:119-26, 1998) proposed a modified approach to constructing confidence intervals for a proportion. A (1 – α) ∙ 100% confidence interval for p is given by

Lower bound:

Upper bound:

where  (x is the number of successes in n trials).

Use this result to answer Problem 1-2.

Problem 1

Cauliflower? Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a 95% confidence interval for the proportion of students who eat cauliflower on Jane’s campus using Agresti and Coull’s method.

Problem 2

Simulation - When Model Requirements Fail A Bernoulli random variable is a variable that is either 0 (a failure) or 1 (a success). The probability of success is denoted p.

(a) Use StatCrunch, MINITAB, or some other statistical spreadsheet to generate 1000 Bernoulli samples of size n = 20 with p = 0.15.

(b) Estimate the population proportion for each of the 1000 Bernoulli samples.

(c) Draw a histogram of the 1000 proportions from part (b). What is the shape of the histogram?

(d) Construct a 95% confidence interval for each of the 1000 Bernoulli samples using the normal model.

(e) What proportion of the intervals do you expect to include the population proportion, p? What proportion of the intervals actually captures the population proportion? Explain any differences.

Questions & Answers

QUESTION:

Problem 43AYU

To deal with issues such as the distribution of p not following a normal distribution A. Agresti and B. Coull (Approximate Is Better Than “Exact” for Interval Estimation of Binomial Proportion. American Statistician, 52:119-26, 1998) proposed a modified approach to constructing confidence intervals for a proportion. A (1 – α) ∙ 100% confidence interval for p is given by

Lower bound:

Upper bound:

where  (x is the number of successes in n trials).

Use this result to answer Problem 1-2.

Problem 1

Cauliflower? Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a 95% confidence interval for the proportion of students who eat cauliflower on Jane’s campus using Agresti and Coull’s method.

Problem 2

Simulation - When Model Requirements Fail A Bernoulli random variable is a variable that is either 0 (a failure) or 1 (a success). The probability of success is denoted p.

(a) Use StatCrunch, MINITAB, or some other statistical spreadsheet to generate 1000 Bernoulli samples of size n = 20 with p = 0.15.

(b) Estimate the population proportion for each of the 1000 Bernoulli samples.

(c) Draw a histogram of the 1000 proportions from part (b). What is the shape of the histogram?

(d) Construct a 95% confidence interval for each of the 1000 Bernoulli samples using the normal model.

(e) What proportion of the intervals do you expect to include the population proportion, p? What proportion of the intervals actually captures the population proportion? Explain any differences.

ANSWER:

Answer :

Step 1:

       For the given information,Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower.

   

               Here given values are x = 2 and n = 20.

              Now,

                     The sample proportion is

                           =

       Now we have to interpret a 95% confidence interval for the proportion of students who eat cauliflower on Jane’s campus using Agresti and Coull’s method.

       

 The formula for confidence interval of the population proportion is

             

          0.16671.96(                  (= 1.96 from standard normal table)

          0.16671.96(0.076)

          0.1667 0.1489

        (0.018 , 0.316)

Lower bound is 0.018 and upper bound is 0.316.

Problem 2

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