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Chapter 7: Problem 2

For each of the following sets of sample outcomes, construct the \(99 \%\) confidence interval for estimating \(P_{u}\). a. \(\begin{aligned} P_{s} &=0.40 \\ N &=548 \end{aligned}\) b. \(\begin{aligned} P_{s} &=0.37 \\ N &=522 \end{aligned}\) c. \(\begin{aligned} P_{s} &=0.79 \\ N &=121 \end{aligned}\) d. \(\begin{aligned} P_{s} &=0.14 \\ N &=100 \end{aligned}\) e. \(\begin{aligned} P_{s} &=0.43 \\ N &=1049 \end{aligned}\) f. \(\begin{aligned} P_{s} &=0.63 \\ N &=300 \end{aligned}\)

Short Answer

Expert verified
a) (0.346, 0.454) b) (0.317, 0.423) c) (0.696, 0.884) d) (0.072, 0.208) e) (0.391, 0.469) f) (0.558, 0.702)

Step by step solution

01

- Understand the formula for confidence interval

The formula for the confidence interval for a population proportion \(\text{P}_{u}\) using a sample proportion \(\text{P}_{s}\) is given by: \[ \text{CI} = \text{P}_{s} \pm Z_{\alpha/2} \sqrt{ \frac{ \text{P}_{s}(1-\text{P}_{s}) } {N} }\] where \(Z_{\alpha/2}\) is the z-score corresponding to the desired confidence level, and \(N\) is the sample size.
02

- Determine the Z-score for a 99% confidence level

For a 99% confidence interval, \(Z_{\alpha/2} = 2.576\). This value is obtained from standard normal distribution tables.
03

- Apply the formula to each set of values

Calculate the confidence interval for each part by substituting the given \( \text{P}_{s} \) and \( N \) values into the confidence interval formula.
04

- Part a

\(\text{P}_{s} = 0.40\), \(N = 548\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.40 \pm 2.576 \sqrt{ \frac{0.40(1-0.40)}{548}} \approx 0.40 \pm 0.054 \Rightarrow (0.346, 0.454)\]
05

- Part b

\(\text{P}_{s} = 0.37\), \(N = 522\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.37 \pm 2.576 \sqrt{ \frac{0.37(1-0.37)}{522}} \approx 0.37 \pm 0.053 \Rightarrow (0.317, 0.423)\]
06

- Part c

\(\text{P}_{s} = 0.79\), \(N = 121\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.79 \pm 2.576 \sqrt{ \frac{0.79(1-0.79)}{121}} \approx 0.79 \pm 0.094 \Rightarrow (0.696, 0.884)\]
07

- Part d

\(\text{P}_{s} = 0.14\), \(N = 100\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.14 \pm 2.576 \sqrt{ \frac{0.14(1-0.14)}{100}} \approx 0.14 \pm 0.068 \Rightarrow (0.072, 0.208)\]
08

- Part e

\(\text{P}_{s} = 0.43\), \(N = 1049\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.43 \pm 2.576 \sqrt{ \frac{0.43(1-0.43)}{1049}} \approx 0.43 \pm 0.039 \Rightarrow (0.391, 0.469)\]
09

- Part f

\(\text{P}_{s} = 0.63\), \(N = 300\) Plugging in the values, the confidence interval is: \[ \text{CI} = 0.63 \pm 2.576 \sqrt{ \frac{0.63(1-0.63)}{300}} \approx 0.63 \pm 0.072 \Rightarrow (0.558, 0.702)\]

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Key Concepts

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

confidence interval
A confidence interval provides a range of values that is likely to contain a population parameter with a certain level of confidence. For example, a 99% confidence interval suggests that if we were to take 100 different samples and compute the interval for each sample, 99 of those intervals would contain the population parameter. The formula for the confidence interval for a population proportion (P_u) using a sample proportion (P_s) is:

\[ \text{CI} = \text{P}_{s} \pm Z_{\alpha/2} \sqrt{ \text{P}_{s}(1-\text{P}_{s}) }/N \]
Here, \( Z_{\alpha/2} \) is the z-score corresponding to the desired confidence level, and \( N \) is the sample size. This formula estimates the range within which the true population proportion is expected to be, based on the observed sample.
population proportion
The population proportion (P_u) is the fraction of the population that has a particular characteristic. For example, if we are interested in the proportion of people that support a particular policy in a town, the population proportion would be the actual fraction of all residents that support the policy. When we can't survey the entire population, we often use a sample proportion (P_s) which is the fraction of the sampled population that has the characteristic. For example, if 40% of a sample of 548 individuals support the policy, then \(P_s = 0.40\).
sample size
Sample size (N) is the number of observations or data points in the sample. It is a crucial factor in statistical analysis. Larger sample sizes generally lead to more precise estimates of the population parameter because they reduce the margin of error in the confidence interval. In our problems, the sample sizes vary, like 548 in part (a) and 100 in part (d). A larger N will typically produce a narrower confidence interval, indicating a more precise estimate.
z-score
The z-score (\(Z_{\alpha/2}\)) represents the number of standard deviations a data point is from the mean in a standard normal distribution. For a 99% confidence level, the z-score is 2.576. This means that we expect the true population proportion to fall within 2.576 standard deviations from the sample proportion 99% of the time. The z-score is a critical value that helps us determine the range or margin of error in our confidence interval.
standard normal distribution
The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is a symmetrical, bell-shaped distribution that allows us to use z-scores to find probabilities and critical values. Most statistical tables and calculations, including our confidence interval calculations, use the standard normal distribution to determine z-scores. It simplifies the process of finding the margin of error for different confidence levels.

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Most popular questions from this chapter

A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to estimate the population mean. Use the \(90 \%\) level. a. An average of \(2.3\) people resides in each household. Standard deviation is \(0.35\). b. There was an average of \(2.1\) television sets \((s=0.10)\) and \(0.78\) telephones \((s=0.55)\) per household. c. The households averaged \(6.0\) hours of television viewing per day \((s=3.0)\)

Two individuals are running for mayor of Shinbone. You conduct an election survey a week before the election and find that \(51 \%\) of the respondents prefer candidate A. Can you predict a winner? Use the \(99 \%\) level. (HINT: In a two-candidate race, what percentage of the vote would the winner need? Does the confidence interval indicate that candidate \(A\) has a sure margin of victory? Remember that while the population parameter is probably \(\alpha=0.01\) in the confidence interval, it may be anywhere in the interval.) $$ \begin{aligned} &P_{s}=0.51 \\ &N=578 \end{aligned} $$

You have developed a series of questions to measure job satisfaction of bus drivers in New York City. A random sample of 100 drivers has an average score of \(10.6\), with a standard deviation of \(2.8\). What is your estimate of the average job satisfaction score for the population as a whole? Use the \(95 \%\) confidence level.

A random sample of 100 inmates at a maximum security prison shows that exactly 10 of the respondents had been the victims of violent crime during their incarceration. Estimate the proportion of victims for the population as a whole, using the \(90 \%\) confidence level. (HINT: Calculate the sample proportion \(P_{s}\) before using Formula 7.3. Remember that proportions are equal to frequency divided by \(N\).)

The results listed next are from a survey given to a random sample of the American public. For each sample statistic, construct a confidence interval estimate of the population parameter at the \(95 \%\) confidence level. Sample size \((N)\) is 2987 throughout. a. The average occupational prestige score was 43.87, with a standard deviation of \(13.52\). b. The respondents reported watching an average of \(2.86\) hours of TV per day, with a standard deviation of \(2.20\). c. The average number of children was \(1.81\), with a standard deviation of \(1.67\). d. Of the 2987 respondents, 876 identified themselves as Catholic. e. Five hundred thirty-five of the respondents said that they had never married. f. The proportion of respondents who said they voted for Bush in the 2004 presidential election was \(0.52\). g. When asked about capital punishment, 2425 of the respondents said that they favored the death penalty for murder.
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