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(e) Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. Lower bound:, Upper bound:O (Round to three decimal places as needed.)

(e) Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. Lower bound:, Upper bound:O (Round to three decimal places as needed.)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question #6

Answer Part E only!!

**Standard Normal Distribution Table** **Page 1:** This section of the Standard Normal Distribution table provides the cumulative probability associated with a given z-score. The table helps you find the probability that a standard normal random variable is less than or equal to z, which is represented as the shaded area under the curve to the left of z in the graph. The graph depicts a standard normal distribution curve with the shaded region indicating the probability area for a specific z-score. The table columns are organized as follows: - The first column shows the z-scores starting from -3.4 up to -0.4, in increments of 0.1. - Each subsequent column (from 0.01 to 0.09) allows you to find more precise z-scores by adding these to the base z-score from the first column. For example: - For a z-score of -1.25, locate the row for -1.2 and then find the column under 0.05. The value at this intersection represents the cumulative probability. **Page 2:** This continuation of the Standard Normal Distribution table provides cumulative probabilities for z-scores ranging from 0.0 to 3.4. The graph is similar to the one on page 1, showing a standard normal distribution with the area to the left of a specific z-score shaded. The table structure is as follows: - The first column shows the z-scores from 0.0 to 3.4. - The subsequent columns (0.00 to 0.09) provide more precision for the z-scores. For instance: - For a z-score of 2.33, find the row for 2.3 and then locate the column under 0.03. The intersection value gives the cumulative probability. This table is an essential tool for statistical analysis, particularly when dealing with questions about the probability of occurrences in a normally distributed dataset. **Note:** To use the tables, identify the whole number and the first decimal in the leftmost column, and the second decimal in the top row. Find the intersection to get the probability.
Transcribed Image Text:**Standard Normal Distribution Table** **Page 1:** This section of the Standard Normal Distribution table provides the cumulative probability associated with a given z-score. The table helps you find the probability that a standard normal random variable is less than or equal to z, which is represented as the shaded area under the curve to the left of z in the graph. The graph depicts a standard normal distribution curve with the shaded region indicating the probability area for a specific z-score. The table columns are organized as follows: - The first column shows the z-scores starting from -3.4 up to -0.4, in increments of 0.1. - Each subsequent column (from 0.01 to 0.09) allows you to find more precise z-scores by adding these to the base z-score from the first column. For example: - For a z-score of -1.25, locate the row for -1.2 and then find the column under 0.05. The value at this intersection represents the cumulative probability. **Page 2:** This continuation of the Standard Normal Distribution table provides cumulative probabilities for z-scores ranging from 0.0 to 3.4. The graph is similar to the one on page 1, showing a standard normal distribution with the area to the left of a specific z-score shaded. The table structure is as follows: - The first column shows the z-scores from 0.0 to 3.4. - The subsequent columns (0.00 to 0.09) provide more precision for the z-scores. For instance: - For a z-score of 2.33, find the row for 2.3 and then locate the column under 0.03. The intersection value gives the cumulative probability. This table is an essential tool for statistical analysis, particularly when dealing with questions about the probability of occurrences in a normally distributed dataset. **Note:** To use the tables, identify the whole number and the first decimal in the leftmost column, and the second decimal in the top row. Find the intersection to get the probability.
A random sample of 1026 adults in a certain large country was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1026 adults surveyed, 515 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. **Click here to view the standard normal distribution table (page 1).** **Click here to view the standard normal distribution table (page 2).** --- **(a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without.** \(\hat{p} = 0.502\) (Round to three decimal places as needed.) --- **(b) Verify that the requirements for constructing a confidence interval about \( p \) are satisfied.** The sample is stated to be a simple random sample, the value of \( n\hat{p}(1-\hat{p}) \) is 256.496, which is greater than or equal to 10, and the sample size can be assumed to be less than or equal to 5% of the population size. (Round to three decimal places as needed.) --- **(c) Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice.** (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) - **A.** We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.471 and 0.533. - **B.** There is a % probability the proportion of adults in the country who believe that televisions are a luxury they could do without is between _ and _. --- **(d) Is it possible that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likely?** It is possible, but not likely that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence interval does not contain 0.60. (Type an integer or a decimal. Do not round.) --- **(e) Use the results of part (c) to construct
Transcribed Image Text:A random sample of 1026 adults in a certain large country was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1026 adults surveyed, 515 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. **Click here to view the standard normal distribution table (page 1).** **Click here to view the standard normal distribution table (page 2).** --- **(a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without.** \(\hat{p} = 0.502\) (Round to three decimal places as needed.) --- **(b) Verify that the requirements for constructing a confidence interval about \( p \) are satisfied.** The sample is stated to be a simple random sample, the value of \( n\hat{p}(1-\hat{p}) \) is 256.496, which is greater than or equal to 10, and the sample size can be assumed to be less than or equal to 5% of the population size. (Round to three decimal places as needed.) --- **(c) Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice.** (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) - **A.** We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.471 and 0.533. - **B.** There is a % probability the proportion of adults in the country who believe that televisions are a luxury they could do without is between _ and _. --- **(d) Is it possible that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likely?** It is possible, but not likely that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence interval does not contain 0.60. (Type an integer or a decimal. Do not round.) --- **(e) Use the results of part (c) to construct
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