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In January 2010 a Gallup Poll asked a random sample of adults, "In general, are you satisfied or dissatisfied with the way things are going in the United States at this time?" In all, 256 said that they were satisfied and the remaining 769 said they were not. Construct and interpret a \(90 \%\) confidence interval for the proportion of adults who are satisfied with how things are going.

Short Answer

Expert verified
The 90% confidence interval is approximately (0.2275, 0.2725).

Step by step solution

01

Identify the Sample Proportion

First, determine the sample proportion (\(\hat{p}\)) of adults who are satisfied. The total number surveyed is \(256 + 769 = 1025\). The sample proportion is the number of satisfied individuals, \(256\), divided by the total \(1025\). Thus, \(\hat{p} = \frac{256}{1025}\).
02

Calculate the Sample Proportion

Calculate the sample proportion:\[ \hat{p} = \frac{256}{1025} \approx 0.25 \]
03

Find the Standard Error

The standard error (SE) is calculated using the formula: \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(n\) is the total sample size. Substitute \(\hat{p} = 0.25\) and \(n = 1025\) into the formula to find the SE.
04

Compute the Standard Error

Substitute the known values into the standard error formula:\[ SE = \sqrt{\frac{0.25(1-0.25)}{1025}} \approx 0.0137 \]
05

Determine the Critical Value

For a \(90\%\) confidence interval, use a Z-score corresponding to \(90\%\), which is \(1.645\). This is the critical value for a normal distribution.
06

Calculate the Margin of Error

The margin of error (ME) is found by multiplying the standard error by the critical value: \(ME = Z \times SE\). Substitute \(Z = 1.645\) and \(SE = 0.0137\) into the formula.
07

Compute the Margin of Error

Calculate the margin of error:\[ ME = 1.645 \times 0.0137 \approx 0.0225 \]
08

Construct the Confidence Interval

The confidence interval is given by: \(\hat{p} \pm ME\). Substitute the values to find \(0.25 \pm 0.0225\).
09

Final Confidence Interval Calculation

Compute the interval: the lower bound is \(0.25 - 0.0225 = 0.2275\) and the upper bound is \(0.25 + 0.0225 = 0.2725\).
10

Interpret the Confidence Interval

The \(90\%\) confidence interval for the proportion of adults who are satisfied is \((0.2275, 0.2725)\). We are \(90\%\) confident that the true proportion of all adults who are satisfied lies within this interval.

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Key Concepts

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

Sample Proportion
The sample proportion is denoted as \(\hat{p}\) and it provides an estimate of the proportion of a population with a particular characteristic, based on a sample. In our context, it represents the proportion of adults who are satisfied with the current state of affairs in the United States.
To calculate the sample proportion, divide the number of individuals with the desired characteristic by the total number of individuals in the sample. This simple formula \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of successes (or satisfied individuals) and \(n\) is the total sample size, gives us \(\hat{p}\).
In this example, \(x = 256\) and \(n = 1025\), so \(\hat{p} = \frac{256}{1025} \approx 0.25\).
  • The sample proportion indicates that 25% of surveyed adults are satisfied.
  • It acts as a representative indicator for the entire population, from which responses were sampled.
Standard Error
The standard error (SE) measures the variability or spread of the sample proportion in repeated sampling. It helps us understand how much the sample proportion \(\hat{p}\) might fluctuate from the true population proportion if we took multiple samples.
To compute the standard error for a proportion, use the formula:
\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Here, \(\hat{p}\) is the sample proportion and \(n\) is the total sample size. This formula reflects the natural diversity in the population proffered through sampling odds.
In our case, inserting \(\hat{p} = 0.25\) and \(n = 1025\) into the formula gives:
\[ SE = \sqrt{\frac{0.25 \times (1 - 0.25)}{1025}} \approx 0.0137 \]
  • In essence, the smaller the standard error, the more precise our sample's estimate is of the actual population parameter.
  • A smaller standard error indicates less variation and suggests that our sample proportion is close to the true population proportion.
Margin of Error
The margin of error quantifies the range within which we expect the true population proportion to fall, with a certain level of confidence. It represents the uncertainty inherent in sampling due to random errors.
To calculate the margin of error (ME), multiply the standard error by the critical value associated with your confidence level. With a 90% confidence level, the critical value (Z-score) is approximately 1.645.
The formula for the margin of error is:
\[ ME = Z \times SE \]
Using our example, where \(SE = 0.0137\), we find:
\[ ME = 1.645 \times 0.0137 \approx 0.0225 \]
  • The margin of error of 0.0225 suggests that we can be 90% confident that the actual proportion differs from our sample proportion by no more than approximately 2.25 percentage points.
  • The smaller the margin of error, the more precise our confidence interval.
Z-Score
A Z-score, or standard score, is a statistical measurement that describes a value's relation to the mean of a group of values. In a standard normal distribution, it tells us how many standard deviations an element is from the mean.
For constructing confidence intervals, the Z-score is crucial because it determines the number of standard deviations to stretch out to capture the desired confidence level's interval.
Each confidence level has a corresponding Z-score. For example:
  • For 90% confidence, the Z-score is 1.645.
  • This means we're choosing a span that covers 1.645 standard deviations away from the mean in both directions.
Using the Z-score, we can calculate the margin of error and thus construct the confidence interval.
In our Gallup poll example, the Z-score helps determine how wide our interval needs to be so that we assert with 90% confidence that the true proportion falls inside it. The Z-score thus serves as a bridge between the standard error and the accuracy of our predictions linked with the entire population.

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Most popular questions from this chapter

Statistics class in Exercise 1 also asked an SRS of 20 boys at their school how many pairs of shoes they have. A \(95 \%\) confidence interval for the difference in the population means (girls - boys) is 10.9 to \(26.5 .\) Interpret the confidence interval and the confidence level.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). Latoya wants to estimate what proportion of the seniors at her boarding high school like the cafeteria food. She interviews an SRS of 50 of the 175 seniors living in the dormitory. She finds that 14 think the cafeteria food is good.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). The small round holes you often see in sea shells were drilled by other sea creatures, who ate the former dwellers of the shells. Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Researchers collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel. \({ }^{11}\) The researchers want to estimate the proportion \(p\) of Oregon whelks that will spontaneously drill into mussels.

A $$ 95 \%$$ confidence interval for the mean body mass index (BMI) of young American women is \(26.8 \pm 0.6\). Discuss whether each of the following explanations is correct. (a) We are confident that \(95 \%\) of all young women have BMI between 26.2 and 27.4 . (b) We are \(95 \%\) confident that future samples of young women will have mean BMI between 26.2 and 27.4 . (c) Any value from 26.2 to 27.4 is believable as the true mean BMI of young American women. (d) If we take many samples, the population mean BMI will be between 26.2 and 27.4 in about \(95 \%\) of those samples. (e) The mean BMI of young American women cannot be 28 .

You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a \(98 \%\) confidence interval for the mean of the population? (a) 2.177 (b) 2.183 (c) 2.326 (d) 2.500 (e) 2.508

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