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Chapter 9: Problem 6

Construct the appropriate confidence interval. A simple random sample of size \(n=785\) adults was asked if they follow college football. Of the 785 surveyed, 275 responded that they did follow college football. Construct a \(95 \%\) confidence interval for the population proportion of adults who follow college football.

Short Answer

Expert verified
\[0.307 \leq p \leq 0.388\]

Step by step solution

01

Identify the sample proportion

Calculate the sample proportion, denoted as \(\hat{p}\). This is given by the number of adults who follow college football divided by the total number of surveyed adults. \(\hat{p} = \frac{275}{785} \).
02

Calculate the standard error

The standard error (SE) of the sample proportion is calculated using the formula: \(SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \).
03

Determine the critical value

For a 95% confidence interval, the critical value (z-score) is approximately \(1.96\). This can be found using standard normal distribution tables.
04

Compute the margin of error

The margin of error (ME) is calculated by multiplying the critical value by the standard error: \(ME = z \cdot SE\).
05

Construct the confidence interval

The confidence interval is given by: \(\hat{p} \pm ME\). This interval provides the range within which the true population proportion is expected to lie with 95% confidence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
To begin constructing a confidence interval, we first need to determine the sample proportion. The sample proportion, denoted as \(\backslash\backslash\hat{p}\backslash\backslash\), is calculated by dividing the number of people who responded positively to the survey by the total number of respondents. In our example, 275 out of 785 adults follow college football, giving us \(\backslash\backslash\hat{p} = \backslash\backslash\frac{275}{785} = 0.3503\backslash\backslash\). This means that around 35.03% of the sampled adults follow college football. This proportion serves as an estimate of the population proportion.
Standard Error
The next step involves calculating the standard error (SE) of the sample proportion. The standard error measures the variability or uncertainty of the sample proportion. To compute the standard error, use the formula: \(\backslash\backslash\ SE = \backslash\backslash\sqrt{ \backslash\backslash\frac{\backslash\backslash\hat{p}(1-\backslash\backslash\hat{p})}{n} } = \backslash\backslash\sqrt{\backslash\backslash\frac{0.3503 \backslash\backslash\times (1-0.3503)}{785}} = 0.0171\backslash\backslash\). This calculation shows that the standard deviation, or uncertainty, of our sample proportion is about 1.71%.
Critical Value
To construct a confidence interval, we also need the critical value. The critical value is determined by the desired confidence level. For a 95% confidence interval, the critical value (z-score) is 1.96. This z-score corresponds to the point on the standard normal distribution where the cumulative probability is 0.975 (since 95% confidence level means 2.5% in each tail of the distribution).
Margin of Error
The margin of error (ME) combines the standard error with the critical value to give us the range of the sample proportion estimate. It is computed by multiplying the critical value by the standard error: \(\backslash\backslash\ ME = z \backslash\backslash\backslash\times SE = 1.96 \backslash\backslash\backslash\times 0.0171 = 0.0335\backslash\backslash\). Thus, our margin of error is approximately 3.35%. This means that the sample proportion might vary by this amount in either direction.
Population Proportion
Finally, we use the sample proportion and the margin of error to find the confidence interval for the population proportion. The confidence interval is calculated as: \(\backslash\backslash\ \backslash\backslash\hat{p} \backslash\backslash \backslash\backslash\backslash\pm ME = 0.3503 \backslash\backslash \backslash\backslash\backslash\pm 0.0335\backslash\backslash\). This gives us an interval of \(\backslash\backslash\ [0.3168, 0.3838]\backslash\backslash\). Therefore, we can be 95% confident that the true population proportion of adults who follow college football falls between 31.68% and 38.38%.

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Most popular questions from this chapter

Blood Alcohol Concentration A random sample of 51 fatal crashes in 2013 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 gram per deciliter \((\mathrm{g} / \mathrm{dL})\) with a standard deviation of \(0.010 \mathrm{~g} / \mathrm{dL}\) (a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive \(\mathrm{BAC}\) (b) In \(2013,\) there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(90 \%\) confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. (d) All 50 states and the District of Columbia use a BAC of \(0.08 \mathrm{~g} / \mathrm{dL}\) as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? Explain.

Determine the point estimate of the population mean and margin of error for each confidence interval. Lower bound: \(20,\) upper bound: 30

Construct the appropriate confidence interval. A simple random sample of size \(n=12\) is drawn from a population that is normally distributed. The sample mean is found to be \(\bar{x}=45,\) and the sample standard deviation is found to be \(s=14\). Construct a \(90 \%\) confidence interval for the population mean.

Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity of the projectile was measured. The sample had a mean muzzle velocity of 863 meters per second and a standard deviation of 2.7 meters per second. Construct and interpret a \(99 \%\) confidence interval for the mean muzzle velocity.

The trade volume of a stock is the number of shares traded on a given day. The following data, in millions (so that 6.16 represents 6,160,000 shares traded), represent the volume of PepsiCo stock traded for a random sample of 40 trading days in 2014. \begin{array}{llllllll} \hline 6.16 & 6.39 & 5.05 & 4.41 & 4.16 & 4.00 & 2.37 & 7.71 \\ \hline 4.98 & 4.02 & 4.95 & 4.97 & 7.54 & 6.22 & 4.84 & 7.29 \\ \hline 5.55 & 4.35 & 4.42 & 5.07 & 8.88 & 4.64 & 4.13 & 3.94 \\ \hline 4.28 & 6.69 & 3.25 & 4.80 & 7.56 & 6.96 & 6.67 & 5.04 \\ \hline 7.28 & 5.32 & 4.92 & 6.92 & 6.10 & 6.71 & 6.23 & 2.42 \\ \hline \end{array} (a) Use the data to compute a point estimate for the population mean number of shares traded per day in 2014 (b) Construct a \(95 \%\) confidence interval for the population mean number of shares traded per day in 2014 . Interpret the confidence interval. (c) A second random sample of 40 days in 2014 resulted in the data shown next. Construct another \(95 \%\) confidence interval for the population mean number of shares traded per day in 2014\. Interpret the confidence interval. $$ \begin{array}{llllrlll} \hline 6.12 & 5.73 & 6.85 & 5.00 & 4.89 & 3.79 & 5.75 & 6.04 \\ \hline 4.49 & 6.34 & 5.90 & 5.44 & 10.96 & 4.54 & 5.46 & 6.58 \\ \hline 8.57 & 3.65 & 4.52 & 7.76 & 5.27 & 4.85 & 4.81 & 6.74 \\ \hline 3.65 & 4.80 & 3.39 & 5.99 & 7.65 & 8.13 & 6.69 & 4.37 \\ \hline 6.89 & 5.08 & 8.37 & 5.68 & 4.96 & 5.14 & 7.84 & 3.71 \\ \hline \end{array} $$ (d) Explain why the confidence intervals obtained in parts (b) and (c) are different.
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