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Chapter 7: Problem 5

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 \((s=3.0)\)

Short Answer

Expert verified
a. [2.257, 2.343], b. [2.088, 2.112] (TV sets) and [0.712, 0.848] (telephones), c. [5.631, 6.369].

Step by step solution

01

- Identify the problem

The task is to construct confidence intervals to estimate the population mean for different variables provided in the sample data: number of people per household, number of television sets and telephones per household, and hours of television viewing per day, all with a 90% confidence level.
02

- Find critical value for 90% confidence level

Use a Z-table or a statistical calculator to find the critical value for a 90% confidence interval. The critical value (Z) for a 90% confidence interval is approximately 1.645.
03

- Calculate margin of error for number of people per household

Compute the margin of error using the formula \(\text{Margin of Error} = Z \times \frac{s}{\sqrt{n}}\), where Z is the critical value, s is the standard deviation, and n is the sample size. For this problem: \(\text{Margin of Error} = 1.645 \times \frac{0.35}{\sqrt{178}} = 0.043\)
04

- Calculate confidence interval for number of people per household

Use the margin of error to calculate the confidence interval. The confidence interval is given by \(\text{Mean} \pm \text{Margin of Error}\). For this problem: Confidence Interval = 2.3 \pm 0.043, which is \([2.257, 2.343]\).
05

- Calculate margin of error for number of television sets per household

For the number of television sets per household: \(\text{Margin of Error} = 1.645 \times \frac{0.10}{\text{\textbackslash sqrt{178}}} = 0.012\)
06

- Calculate confidence interval for number of television sets per household

Confidence Interval for television sets: \(\text{Mean} \pm \text{Margin of Error} = 2.1 \pm 0.012\). This gives \([2.088, 2.112]\).
07

- Calculate margin of error for number of telephones per household

For the number of telephones per household: \(\text{Margin of Error} = 1.645 \times \frac{0.55}{\text{\textbackslash sqrt{178}}} = 0.068\)
08

- Calculate confidence interval for number of telephones per household

Confidence Interval for telephones: \(\text{Mean} \pm \text{Margin of Error} = 0.78 \pm 0.068\). This gives \([0.712, 0.848]\).
09

- Calculate margin of error for hours of television viewing per day

For the hours of television viewing per day: \(\text{Margin of Error} = 1.645 \times \frac{3.0}{\text{\textbackslash sqrt{178}}} = 0.369\)
10

- Calculate confidence interval for hours of television viewing per day

Confidence Interval for television viewing: \(\text{Mean} \pm \text{Margin of Error} = 6.0 \pm 0.369\). This gives \([5.631, 6.369]\).

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Key Concepts

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

population mean estimation
Population mean estimation is a fundamental concept in statistics. It involves using data from a sample to make inferences about the mean (average) of a larger population. In our exercise, the sample includes information from 178 households about various variables, like the number of people per household and hours of TV viewing. The goal is to make an educated guess about these averages for the entire population based on our sample data.
90% confidence level
A 90% confidence level indicates that we are 90% confident that the true population mean lies within our calculated interval. In simpler terms, if we were to take 100 different samples and build a confidence interval from each, approximately 90 of those intervals would contain the true population mean. This confidence level is slightly more conservative than the commonly used 95% level, giving a narrower range but still providing reasonable assurance. The critical value for a 90% confidence level, which is used in our calculations, is 1.645.
margin of error calculation
The margin of error quantifies the amount of random sampling error in our estimates. It's calculated using the formula: the formula \(\text{Margin of Error} = Z \times \frac{s}{\text{\text{\backslash}}sqrt{n}}\), where 'Z' is the critical value, 's' is the standard deviation, and 'n' is the sample size. The Z-value depends on our chosen confidence level. For a 90% confidence level, it's 1.645. The margin of error helps us create the range that forms our confidence interval. For instance, in our exercise, we calculated a margin of error of 0.043 for the number of people per household and 0.369 for daily TV viewing hours.
Z-table usage
The Z-table is a crucial tool in statistics for finding the critical values needed in confidence interval calculations. It shows the area (probability) under the standard normal curve corresponding to different Z-values. For a 90% confidence interval, the critical value 'Z' is 1.645. This value corresponds to a cumulative probability of 0.95 (covering the middle 90% of the distribution). Using a Z-table, or a statistical calculator, we can quickly lookup these critical values to use in our margin of error calculations. Proper Z-table usage ensures that we accurately build our confidence intervals, providing valid and reliable estimates of the population mean.

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Most popular questions from this chapter

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 fraternities and sororities at St. Algebra College have been plagued by declining membership over the past several years and want to know if the incoming freshman class will be a fertile recruiting ground. Not having enough money to survey all 1600 freshmen, they commission you to survey the interests of a random sample. You find that 35 of your 150 respondents are "extremely" interested in social clubs. At the \(95 \%\) level, what is your estimate of the number of freshmen who would be extremely interested? (HINT: The high and low values of your final confidence interval are proportions. How can proportions also be expressed as numbers?)

A random sample of } 260 \text { workers in a high-rise }\end{array}\( office building revealed that \)30 \%\( were very satisfied with the quality of elevator service. At the \)99 \%$ level, what is your estimate of the population value?

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.

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. \(\begin{aligned} P_{s} &=0.63 \\ N &=300 \end{aligned}\)
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