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Answer:
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:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
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:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table.
Compute the 95% confidence interval for the population proportion as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=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)[/tex]
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%.
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:
[tex]C.I. = p\ \pm z_\frac{\alpha}{2} \times \sqrt{\frac{p(1-p)}{n} }[/tex]
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.
[tex]z_\frac{\alpha}{2} = z_\frac{0.05}{2} = z_0._0_2_5 = 1.96[/tex]
To calculate the critical value of z for 95% confidence level as follows:
Compute the 95% confidence interval for the population proportion as follows:
[tex]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)[/tex]
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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