Confidence Intervals for Population Mean: Videos & Practice Problems

In this chapter, we explore the construction of confidence intervals for a population mean, denoted as μ. A confidence interval provides a range of values that is likely to contain the true population mean based on sample data. To illustrate this process, we will work through an example involving travel times.

Consider a scenario where a sample of 36 trips to work yields a sample mean (x̄) of 1 hour (or 60 minutes), with a known population standard deviation (σ) of 18 minutes. To construct a 90% confidence interval for the true population mean travel time, we follow a systematic approach.

First, we verify the conditions necessary for constructing the confidence interval. We assume the sample is random unless stated otherwise, and since our sample size (n) is 36, which is greater than 30, we can proceed with the normal distribution assumption.

Next, we determine the critical z value (z_α/2) for our confidence level. The confidence level (C) is 90%, leading to an alpha (α) of 0.10. Thus, α/2 equals 0.05. Using a z-table or calculator, we find that z_α/2 is approximately 1.645 for this confidence level.

With the critical z value established, we calculate the margin of error (E) using the formula:

$$E = z_{α/2} \times \frac{σ}{\sqrt{n}}$$

Substituting our values, we have:

$$E = 1.645 \times \frac{18}{\sqrt{36}}$$

Calculating this gives us a margin of error of approximately 4.935 minutes.

Finally, we construct the confidence interval by taking the sample mean and adjusting it by the margin of error. The lower bound is calculated as:

$$\text{Lower Bound} = x̄ - E = 60 - 4.935 = 55.065$$

And the upper bound is:

$$\text{Upper Bound} = x̄ + E = 60 + 4.935 = 64.935$$

Thus, the 90% confidence interval for the true population mean travel time is (55.065, 64.935) minutes. This means we are 90% confident that the average travel time to work lies within this range.

Understanding how to construct and interpret confidence intervals is crucial in statistics, as it allows us to make informed conclusions about population parameters based on sample data.

In this problem, we analyze the time it takes to find a parking spot, with a sample mean of 1 hour and 22 minutes (or 82 minutes) and a population standard deviation of 15 minutes, based on a sample of 100 days. Our goal is to construct a 99% confidence interval for the true population mean.

To begin, we confirm that our sample is random and that the sample size is sufficiently large (greater than 30), which allows us to proceed with the analysis. Next, we determine the critical z-value for our confidence level. For a 99% confidence interval, we calculate the alpha level as follows:

$$ \alpha = 1 - 0.99 = 0.01 $$

Then, we find alpha over two:

$$ \frac{\alpha}{2} = \frac{0.01}{2} = 0.005 $$

Using a z-table or calculator, we find the critical z-value, \( z_{\alpha/2} \), to be 2.576.

Next, we calculate the margin of error (E) using the formula:

$$ E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} $$

Substituting the known values:

$$ E = 2.576 \times \frac{15}{\sqrt{100}} $$

Calculating this gives us:

$$ E \approx 3.864 $$

With the margin of error determined, we can construct the confidence interval by subtracting and adding the margin of error to the sample mean:

Lower bound: $$ 82 - 3.864 \approx 78.136 $$

Upper bound: $$ 82 + 3.864 \approx 85.864 $$

Thus, the 99% confidence interval for the true population mean time to find a parking spot is approximately (78.136 minutes, 85.864 minutes).

In interpreting this result, we can confidently state that we are 99% certain that the true mean time it takes to find a parking spot lies within this interval. This statistical analysis provides valuable insight into the expected time drivers may spend searching for parking, which can be crucial for urban planning and traffic management.

When constructing a confidence interval for a population mean (\(\mu\)), the approach varies depending on whether the population standard deviation (\(\sigma\)) is known. If \(\sigma\) is unknown, we can use the sample standard deviation (\(s\)) to estimate it. In this scenario, the sampling distribution of the sample mean (\(\bar{x}\)) follows a t-distribution rather than a normal distribution.

The t-distribution resembles the normal distribution but is shorter and wider, with a mean of zero and a standard deviation greater than one, which varies based on the sample size. A crucial aspect of the t-distribution is the concept of degrees of freedom, calculated as \(n - 1\), where \(n\) is the sample size. As the sample size increases, the degrees of freedom also increase, causing the t-distribution to approach the shape of the normal distribution.

To construct a confidence interval, we need to determine the critical t-value (\(t_{\alpha/2}\)) when \(\sigma\) is unknown. This involves knowing the significance level (\(\alpha\)) and the degrees of freedom. For example, for a 95% confidence interval, \(\alpha/2\) is 0.025. Using a t-table, we can find the critical t-value corresponding to our degrees of freedom. For a sample size of 31, the degrees of freedom would be 30. Looking up \(t_{0.025}\) for 30 degrees of freedom yields a critical t-value of approximately 2.042.

In contrast, if \(\sigma\) is known, we use the critical z-value (\(z_{\alpha/2}\)). For the same 95% confidence interval, the critical z-value is -1.96 (for the left tail), which translates to a positive value of 1.96 for the right tail. It is important to note that the critical t-value is generally greater than the critical z-value for the same confidence level, reflecting the wider margin of error associated with the t-distribution.

Before constructing a confidence interval using either critical value, ensure that the sample is random and that the underlying population is normally distributed or that the sample size is greater than 30. This verification is essential for the validity of the confidence interval.

When constructing a confidence interval for the population mean (\(\mu\)), it is essential to determine whether to use the standard normal distribution or the t distribution. This decision hinges on several factors, including the sample size, whether the population standard deviation (\(\sigma\)) is known, and the normality of the data.

In the first scenario, we have a sample of 18 test scores with a sample mean of 74 and a population standard deviation of 6. Although the sample is random, the sample size is less than 30, which necessitates that the data be normally distributed to proceed. Since we lack information confirming the normality of the data, we cannot construct a confidence interval using either the standard normal or t distribution.

In the second scenario, we measure the heights of 87 men, yielding an average height of 5 feet 10 inches with a population standard deviation of 5 inches. Here, the sample is random, the sample size exceeds 30, and we are told that the heights are normally distributed. Since we know the population standard deviation, we can use the standard normal distribution with a critical value of \(z_{\alpha/2}\) to construct a 99% confidence interval for the population mean height.

In the final scenario, we analyze the spending of 25 randomly selected students, who spend an average of $500 on textbooks with a sample standard deviation of $50. The sample size is again less than 30, but we are informed that the spending is normally distributed. Since we only know the sample standard deviation (\(s\)) and not the population standard deviation, we must use the t distribution with a critical value of \(t_{\alpha/2}\) to construct a 98% confidence interval for the population mean textbook spending.

In summary, the choice between the standard normal and t distributions is influenced by the sample size, the knowledge of the population standard deviation, and the normality of the data. Understanding these factors is crucial for accurately constructing confidence intervals.

To construct a confidence interval for a population mean (\(\mu\)) when the population standard deviation (\(\sigma\)) is unknown, we utilize the sample standard deviation (\(s\)) and the critical t-value (\(t_{\alpha/2}\)). This approach is essential when dealing with smaller sample sizes or when the population standard deviation is not available.

In a practical example, consider a scenario where a sample of 36 trips to work yields a sample mean of 1 hour (or 60 minutes) and a sample standard deviation of 21 minutes. To create a 90% confidence interval for the true population mean travel time, we follow a systematic process.

First, we verify that the sample is random and that either the sample mean is normally distributed or the sample size is greater than 30. In this case, with \(n = 36\), we can proceed confidently.

Next, we determine the critical t-value. The confidence level of 90% implies an alpha level (\(\alpha\)) of 0.10, leading to \(\alpha/2 = 0.05\). The degrees of freedom (df) are calculated as \(n - 1\), which gives us \(35\) for this example. Using a t-distribution table or calculator, we find that \(t_{\alpha/2} = 1.69\).

Now, we calculate the margin of error (\(E\)). The formula for the margin of error when using the t-distribution is:

\[ E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} \]

Substituting the values, we have:

\[ E = 1.69 \times \frac{21}{\sqrt{36}} = 1.69 \times 3.5 = 5.915 \]

With the margin of error calculated, we can find the confidence interval by adding and subtracting the margin of error from the sample mean:

Lower bound: \(60 - 5.915 = 54.085\) minutes

Upper bound: \(60 + 5.915 = 65.915\) minutes

Thus, the 90% confidence interval for the mean travel time is \((54.085, 65.915)\) minutes. This means we can be 90% confident that the true mean travel time to work lies within this interval.

Understanding how to construct and interpret confidence intervals is crucial for statistical analysis, allowing us to make informed conclusions about population parameters based on sample data.

In this example, we explore how to estimate the average number of sick days taken by employees using statistical methods, specifically focusing on constructing a confidence interval for the population mean (μ) based on a sample mean (X̄) and sample standard deviation (s). The initial scenario involves an HR manager who collects data from a random sample of 15 employees, finding a sample mean of 6.8 days and a sample standard deviation of 2.4 days.

To construct a confidence interval, we first verify that the sample is random, which it is. Next, we check the normality of the data distribution. Since the sample size is less than 30, we assume normality unless evidence suggests otherwise. With these conditions met, we proceed to calculate the critical value using the t-distribution, as the population standard deviation (σ) is unknown.

The critical value is determined by calculating α/2, which is 0.05 for a 90% confidence level. The degrees of freedom (df) are calculated as n - 1, resulting in 14 for our sample of 15 employees. Using a t-table, we find the critical t-score for df = 14 at α/2 = 0.05, which is approximately 1.761.

The margin of error (E) is then calculated using the formula: E = t \cdot \frac{s}{\sqrt{n}}. Substituting our values, we find the margin of error to be approximately 1.09. The confidence interval is constructed by subtracting and adding the margin of error to the sample mean: [X̄ - E, X̄ + E] = [6.8 - 1.09, 6.8 + 1.09] = [5.71, 7.89]. This means we can be 90% confident that the true average number of sick days taken by all employees falls within this interval.

In the second scenario, we increase the sample size to 41 employees while keeping the sample mean and standard deviation the same. The degrees of freedom now become 40. The critical t-score for df = 40 at α/2 = 0.05 is approximately 1.684. The margin of error is recalculated using the same formula, resulting in a smaller margin of error of about 0.63 due to the increased sample size.

The new confidence interval is then calculated as [X̄ - E, X̄ + E] = [6.8 - 0.63, 6.8 + 0.63] = [6.17, 7.43]. This demonstrates that as the sample size increases, the margin of error decreases, leading to a more precise estimate of the population mean. Thus, increasing the sample size enhances the reliability of our confidence intervals, allowing for better approximations of the true parameter.