Explanation Calculation: Step-by-step procedure to generate 200 random samples using the MINITAB software: Click on Calc > Random Data > Binomial Distribution . In Number of rows of data to generate section, enter 200 . In the Store in column , enter column C1 named as Random Numbers . For Number of trails , enter 20 . For Probability , enter 0.6 . Click OK .. Output using the MINITAB is given below: Random numbers 9 15 12 16 12 9 15 11 12 10 10 9 14 16 11 11 9 15 13 12 11 6 15 11 15 9 11 11 14 11 9 13 9 12 13 12 13 15 12 13 11 9 10 12 9 12 14 10 15 12 10 8 11 12 10 15 13 10 11 10 14 8 12 14 13 12 10 12 13 7 15 13 11 8 17 11 10 10 15 12 9 15 11 10 13 12 11 10 10 17 12 15 12 12 18 15 9 11 14 9 10 15 13 14 12 9 14 17 15 6 9 13 13 14 13 17 15 15 13 13 15 11 10 9 9 8 14 12 12 13 9 12 10 8 14 13 12 7 15 13 8 9 15 14 11 9 11 14 11 11 14 9 11 15 10 11 14 14 14 11 9 10 11 17 12 13 15 10 12 8 8 13 14 15 9 16 11 15 10 7 11 13 16 11 8 14 9 12 10 12 12 13 11 15 12 12 12 16 12 12 From the Standard Normal Table, the z value for 90% level is 1.645. The general formula for the confidence interval for the binomial parameter is, p ^ ± z 1 − α 2 p ^ q ^ n The lower bound and upper bound is tabulated below: Random Numbers p ^ q ^ = 1 − p ^ z 1 − α 2 S E = z 1 − α 2 p ^ q ^ n Lower bound Upper bound 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 6 0.3 0.7 1.645 0.1686 0.1314 0.4686 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 17 0.85 0.15 1.645 0.1313 0.7187 0.9813 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 11 0

8th Edition

ISBN: 9781305268920

Author: Bernard Rosner

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Addition Rule of Probability

It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.

Expected Value

When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known …

Probability Distributions

Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosi…

Basic Probability

The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the prob…

Videos

Question

Chapter 6, Problem 93P

To determine

Check whether the large-sample binomial confidence limit formula is adequate for the given distribution.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6, Problem 92P

Chapter 6, Problem 94P

Students have asked these similar questions

Example 4 (Part 2) We can use Statkey to take 50 different random samples of size 20 each, find the mean of each sample, and compute a confidence interval for each one. The graph of the sampling distribution of the means is on the left below, and that of the 50 confidence intervals is on the right. 1. What does each dot on the left hand dotplot represent? StatKey Sampling Distribution for a Mean Percent with Internet Access (Countries) ▾ Show Data Table Edit Data Choose samples of size n = 20 Upload File Change Column(s) Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Reset Plot Sampling Dotplot of Mean Left Tail Two-Tail Right Tail 60 50 40 40 30 20 20 10 samples = 50 mean = 41.626 std. error = 5.089 : .: : :: 0 25 30 35 40 45 50 55 60 41.626 Data Plots Confidence Intervals 95%-> Confidence Intervals Coverage 48/50 = 96% 20 40 60 80 2. Circle the confidence intervals that failed to capture the true mean. 3. Circle the sample means that produced those…

Example 4 (Part 1) One of the datasets in the Lock book contains information about 215 countries of the world. One of the variables is the percentage of people in the country who have access to the internet. We have data for 203 of those countries. The plot on the right shows a dotplot of the data. 1. What are the cases? Population n = 203, mean = 43.024 median = 43.5, stdev = 29.259 20 2. What does each dot on the dotplot represent? 15 10 5 20 40 43.024 60 80 3. What type of data is do we collect from the cases, quantitative or categorical?

Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of n =6 professional basketball players gave the following information. x 63 79 70 80 84 87 y 46 49 45 55 57 58 Find Se. Round your answer to three decimal places.

Chapter 6 Solutions

Fundamentals of Biostatistics

Ch. 6.5 - What is the difference between a standard...Ch. 6.5 - What does a 95% CI mean?Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...

Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - A nine-laboratory cooperative study was performed...Ch. 6 - A nine-laboratory cooperative study was performed...Ch. 6 - A nine-laboratory cooperative study was performed...Ch. 6 - A nine-laboratory cooperative study was performed...Ch. 6 - A nine-laboratory cooperative study was performed...Ch. 6 - A study of psychological and physiological health...Ch. 6 - A study of psychological and physiological health...Ch. 6 - A study is conducted to test the hypothesis that...Ch. 6 - A study is conducted to test the hypothesis that...Ch. 6 - Suppose a clinical trial is conducted to test the...Ch. 6 - Suppose a clinical trial is conducted to test the...Ch. 6 - Suppose a clinical trial is conducted to test the...Ch. 6 - Suppose we want to estimate the concentration...Ch. 6 - Suppose we want to estimate the concentration...Ch. 6 - Suppose we want to estimate the concentration...Ch. 6 - Much discussion has taken place concerning...Ch. 6 - Much discussion has taken place concerning...Ch. 6 - Much discussion has taken place concerning...Ch. 6 - A new assay is developed to obtain the...Ch. 6 - A new assay is developed to obtain the...Ch. 6 - A new assay is developed to obtain the...Ch. 6 - A new assay is developed to obtain the...Ch. 6 - Suppose 100 hypertensive people are given an...Ch. 6 - Suppose 100 hypertensive people are given an...Ch. 6 - Suppose 100 hypertensive people are given an...Ch. 6 - Suppose we decide a better measure of the...Ch. 6 - Suppose we decide a better measure of the...Ch. 6 - Suppose we decide a better measure of the...Ch. 6 - Suppose we decide a better measure of the...Ch. 6 - Draw six random samples of size 5 from the data in...Ch. 6 - Draw six random samples of size 5 from the data in...Ch. 6 - Draw six random samples of size 5 from the data in...Ch. 6 - Draw six random samples of size 5 from the data in...Ch. 6 - Draw six random samples of size 5 from the data in...Ch. 6 - Figure 6.4b (p. 172) plotted the sampling...Ch. 6 - Figure 6.4b (p. 172) plotted the sampling...Ch. 6 - Figure 6.4b (p. 172) plotted the sampling...Ch. 6 - Figure 6.4b (p. 172) plotted the sampling...Ch. 6 - The etiology of high blood pressure remains a...Ch. 6 - The etiology of high blood pressure remains a...Ch. 6 - The etiology of high blood pressure remains a...Ch. 6 - Data Set SEXRAT.DAT, at www.cengagebrain.com,...Ch. 6 - A cohort of hemophiliacs is followed to elicit...Ch. 6 - A cohort of hemophiliacs is followed to elicit...Ch. 6 - A cohort of hemophiliacs is followed to elicit...Ch. 6 - We have previously described Data Set LEAD.DAT (at...Ch. 6 - We have previously described Data Set LEAD.DAT (at...Ch. 6 - We have previously described Data Set LEAD.DAT (at...Ch. 6 - Data Set NIFED.DAT (at www.cengagebrain.com) was...Ch. 6 - Data Set NIFED.DAT (at www.cengagebrain.com) was...Ch. 6 - Data Set NIFED.DAT (at www.cengagebrain.com) was...Ch. 6 - Suppose that there are 6 deaths due to bladder...Ch. 6 - Suppose that there are 4 deaths due to stomach...Ch. 6 - The value of mammography as a screening test for...Ch. 6 - The value of mammography as a screening test for...Ch. 6 - The value of mammography as a screening test for...Ch. 6 - The value of mammography as a screening test for...Ch. 6 - On the computer, draw 1000 bootstrap samples of...Ch. 6 - On the computer, draw 1000 bootstrap samples of...Ch. 6 - A study was performed to investigate the...Ch. 6 - A study was performed to investigate the...Ch. 6 - Refer to Data Set BONEDEN.DAT at...Ch. 6 - Refer to Data Set BONEDEN.DAT at...Ch. 6 - Prob. 88P Ch. 6 - Prob. 89P Ch. 6 - Prob. 90P Ch. 6 - Prob. 91P Ch. 6 - Prob. 92P Ch. 6 - Prob. 93P Ch. 6 - Prob. 94P Ch. 6 - Prob. 95P Ch. 6 - Prob. 96P Ch. 6 - A patient who is taking antihypertensive...Ch. 6 - A patient who is taking antihypertensive...Ch. 6 - A patient who is taking antihypertensive...Ch. 6 - A patient who is taking antihypertensive...Ch. 6 - A patient who is taking antihypertensive...Ch. 6 - Injuries are common in football and may be related...Ch. 6 - Injuries are common in football and may be related...Ch. 6 - A hypertensive patient has been on...Ch. 6 - A hypertensive patient has been on...Ch. 6 - A hypertensive patient has been on...Ch. 6 - A hypertensive patient has been on...Ch. 6 - A radiologist investigates whether a new (less...Ch. 6 - A radiologist investigates whether a new (less...Ch. 6 - The estimation of allele probabilities is...Ch. 6 - The estimation of allele probabilities is...Ch. 6 - The estimation of allele probabilities is...Ch. 6 - As an experiment, 10,000 people are completely...Ch. 6 - As an experiment, 10,000 people are completely...Ch. 6 - Which of the following sample results yields a 95%...Ch. 6 - Suppose we observe 30 cases of colon cancer over...

Knowledge Booster

$Statistics$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.