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Chapter 9: Problem 5

Construct a confidence interval of the population proportion at the given level of confidence. $$x=30, n=150,90 \% \text { confidence }$$

Short Answer

Expert verified
The 90% confidence interval for the population proportion is \[ (0.1463, 0.2537) \].

Step by step solution

01

Find the sample proportion

The sample proportion (\(\bar{p}\)) is calculated by dividing the number of successes (x) by the total sample size (n). \[ \bar{p} = \frac{x}{n} = \frac{30}{150} = 0.2 \]
02

Determine the critical value

For a 90% confidence level, find the corresponding z-score. The critical value (z) is the z-score that corresponds to the middle 90% of the standard normal distribution. From standard normal distribution tables, \[ z_{\alpha/2} = 1.645 \]
03

Calculate the standard error

The standard error (SE) is calculated using the formula: \[ SE = \sqrt{ \frac{\bar{p}(1 - \bar{p})}{n} } \] Plug in the values: \[ SE = \sqrt{ \frac{0.2 \times (1 - 0.2)}{150} } = \sqrt{ \frac{0.2 \times 0.8}{150} } = \sqrt{ \frac{0.16}{150} } = \sqrt{0.001067} \approx 0.0327 \]
04

Construct the confidence interval

The confidence interval (CI) is given by \[ \bar{p} \pm z \cdot SE \] Substitute the values: \[ 0.2 \pm 1.645 \cdot 0.0327 \] Calculate the interval: \[ 0.2 - 0.0537 = 0.1463 \] \[ 0.2 + 0.0537 = 0.2537 \] Therefore, the 90% confidence interval for the population proportion is \[ (0.1463, 0.2537) \]

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Key Concepts

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

population proportion
A population proportion refers to the fraction or percentage of the total population that possesses a particular attribute of interest. In our context, it measures how many individuals in a larger group exhibit a specific characteristic compared to the total number of people in that group.
To estimate this proportion precisely using a sample, we calculate the sample proportion. For instance, in our example, the sample proportion (\(\bar{p}\)) is computed by dividing the number of successes (\(x\)) by the total sample size (\(n\)). Here, \(\bar{p} = \frac{30}{150} = 0.2\). This means that 20% of our sample exhibits the trait in question.
standard error
The standard error (SE) helps us understand the variability or uncertainty of the sample proportion. It gives an idea of how much the sample proportion (\(\bar{p}\)) is likely to differ from the true population proportion. A smaller SE indicates more reliable sample proportion estimates.
To calculate the SE, we use the formula: \[ SE = \sqrt{ \frac{\bar{p}(1 - \bar{p})}{n} } \] Plugging in our values, we get: \[ SE = \sqrt{ \frac{0.2 \times (1 - 0.2)}{150} } = \sqrt{ \frac{0.16}{150} } = \sqrt{0.001067} \approx 0.0327 \] This indicates the expected deviation of the sample proportion from the true population proportion.
critical value
A critical value defines the amount of standard deviation away from the mean needed to achieve our desired confidence level. It is denoted as \(z_{\alpha/2}\) and is obtained from standard normal distribution tables.
The critical value accounts for the precision we require in our estimate. For example, for a 90% confidence level, the critical value (z) is 1.645. This value signifies the cut-off points that leave 5% in each tail of the distribution, ensuring our confidence interval covers the middle 90%.
confidence level
The confidence level represents how certain we are that our confidence interval contains the true population proportion. A 90% confidence level means that if we were to take 100 different samples and compute the intervals for each, about 90 of them would contain the true population proportion.
Using our values, the calculation for the confidence interval (\text {CI}) is: \[ \bar{p} \pm z \cdot SE \] This gives us: \[ 0.2 \pm 1.645 \cdot 0.0327 \approx (0.1463, 0.2537) \] Therefore, we are 90% confident that the true population proportion falls within the interval (0.1463, 0.2537).

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Most popular questions from this chapter

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 about the population proportion of adults who follow college football.

IQ scores based on the Wechsler Intelligence Scale for Children (WISC) are known to be normally distributed with \(\mu=100\) and \(\sigma=15\) (a) Simulate obtaining 20 samples of size \(n=15\) from this population. (b) Obtain the sample mean and standard deviation for each of the 20 samples. (c) Construct \(95 \%\) t-intervals for each of the 20 samples. (d) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

Construct a confidence interval of the population proportion at the given level of confidence. $$x=860, n=1100,94 \% \text { confidence }$$

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A school social worker wishes to estimate the mean amount of time each week that high school students spend with friends. She obtains a random sample of 1175 high school students and finds that the mean weekly time spent with friends is 9.2 hours. Assuming the population standard deviation amount of time spent with friends is 6.7 hours, construct and interpret a \(90 \%\) confidence interval for the mean time spent with friends each week.
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