Mathematics, 22.04.2020 03:03 david6835

An article included the following statement: "Few people believe there's much reality in reality TV: a total of 86% said the shows are either 'totally made up' or 'mostly distorted.'" This statement was based on a survey of 1006 randomly selected adults. Compute a bound on the error (based on 95% confidence) of estimation for the reported proportion of 0.86. (Round your answer to three decimal places.) Interpret the bound. (Round your answers to one decimal place.) We are % confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within % of the sample proportion of %.

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95% confident that the proportion of all adults who believe that the shows are either made up" or "mostly distorted" is within 83.9% and 88.1%.

Given that,

An article included the following statement: "Few people believe there's much reality in reality TV:

Total of 86% said the shows are either 'totally made up' or 'mostly distorted.'" This statement was based on a survey of 1006 randomly selected adults

We have to determine,

Compute a bound on the error (based on 95% confidence) of estimation for the reported proportion of 0.86

According to the question,

Let, p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected.

These adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

The (1 - α)% confidence interval for the population proportion is:

$C.I. = p\ \pm z_\frac{\alpha}{2} \times \sqrt{\frac{p(1-p)}{n} }$

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

$z_\frac{\alpha}{2} = z_\frac{0.05}{2} = z_0._0_2_5 = 1.96$

To calculate the critical value of z for 95% confidence level as follows:

Compute the 95% confidence interval for the population proportion as follows:

$C.I. = p \pm z_\frac{\alpha}{2} \sqrt{\frac{p(1-p)}{n} } \\\\C.I. = 0.86\pm 1.96\sqrt{\frac{0.86(1-0.86)}{1006} }\\\\C.I. = 0.86 \pm 0.0214\\\\C.I. = (0.8366, 0.8814)\\\\C.I = (0.839, 0.88)$

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

Hence, 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

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Answer from: living8539

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

Step-by-step explanation:

Let p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected. Of these adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

The (1 - α)% confidence interval for the population proportion is:

$CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.86\pm 1.96\times \sqrt{\frac{0.86(1-0.86)}{1006}}\\=0.86\pm 0.0214\\=(0.8386, 0.8814)\\\approx (0.839, 0.881)$

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

This confidence interval implies that:

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

Answer from: Quest

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Answer from: Quest

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step-by-step explanation:

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