Confidence Intervals for Population Mean: Videos & Practice Problems

In this chapter, we explored the concept of confidence intervals, specifically focusing on constructing a confidence interval for a population mean, denoted as μ. To illustrate this process, we will work through an example involving travel time to work.

Consider a scenario where, over 36 trips to work, the sample mean travel time is found to be 1 hour (or 60 minutes). We aim to construct a 90% confidence interval for the true population mean travel time, using a population standard deviation (σ) of 18 minutes. Since the population standard deviation is provided, we can utilize the sample mean (x̄) as a point estimate and calculate the margin of error (E) using the formula:

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

Before proceeding, we must verify that the conditions for constructing a confidence interval are met. First, we check if the sample is random. In this case, we assume randomness since the problem does not indicate otherwise. Next, we confirm that the sample size (n) is greater than 30, which it is, as n = 36. This allows us to proceed with the construction of the confidence interval.

Next, we need to determine the critical value, \( z_{\alpha/2} \). The value of α is calculated as \( 1 - c \), where c is the confidence level. For a 90% confidence level, α is 0.1, and thus \( \alpha/2 \) is 0.05. Using a z-table or calculator, we find that \( z_{\alpha/2} = 1.645 \).

Now, we can calculate the margin of error:

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

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 mean travel time is from 55.065 minutes to 64.935 minutes. This means we are 90% confident that the true mean travel time to work lies within this interval.

In summary, constructing a confidence interval involves verifying sample conditions, calculating the critical z value, determining the margin of error, and finally establishing the interval itself. This process is essential for making informed statistical inferences about population parameters based on sample data.

In this problem, we analyze the time it takes to find a parking spot, with a 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 and interpret a 99% confidence interval for the true population mean.

First, we confirm that our sample is random and that the sample size is sufficient for applying the Central Limit Theorem. Since our sample size is 100, which is greater than 30, we can proceed with the calculations.

Next, we determine the critical z-value for a 99% confidence interval. The alpha level (α) is calculated as:

α = 1 - 0.99 = 0.01

To find α/2, we divide by 2:

α/2 = 0.01 / 2 = 0.005

Using a z-table or calculator, we find the critical z-value (z_α/2) corresponding to 0.005, which is 2.576.

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

E = z_α/2 × (σ / √n)

Substituting the values, we have:

E = 2.576 × (15 / √100) = 2.576 × 1.5 = 3.864

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

Lower bound = 82 - 3.864 = 78.136

Upper bound = 82 + 3.864 = 85.864

Thus, the 99% confidence interval for the true population mean time to find a parking spot is (78.136, 85.864) minutes. This means we are 99% confident that the actual average time it takes to find a parking spot falls within this range.

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 0 and a standard deviation greater than 1, 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 value can be found using a t-table or calculator, similar to how we find the critical z-value (\(z_{\alpha/2}\)) when \(\sigma\) is known. For example, to find the critical values for a 95% confidence interval with \(\alpha/2 = 0.025\) and a sample size of 31:

1. **Known \(\sigma\)**: We use the critical z-value. From the z-table, the critical z-value corresponding to \(\alpha/2 = 0.025\) is approximately \(-1.96\), leading to a positive critical z-value of \(1.96\).

2. **Unknown \(\sigma\)**: We need to find the critical t-value. With a sample size of 31, the degrees of freedom are \(30\) (i.e., \(31 - 1\)). Referring to the t-table for \(\alpha/2 = 0.025\) and \(30\) degrees of freedom, we find the critical t-value to be approximately \(2.042\).

It is important to note that the critical t-value is always greater than the critical z-value for the same confidence level due to the wider nature of the t-distribution, which accounts for increased uncertainty when estimating the population standard deviation.

Before constructing a confidence interval using these critical values, ensure that the sample is random and that either the population distribution is normal or 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, as each corresponds to different critical values. The choice depends on the sample size, whether the population standard deviation (\(\sigma\)) is known, and the normality of the data.

In the first scenario, we randomly select 18 test scores with a sample mean of 74 and a population standard deviation of 6. Although we have a known population standard deviation, the sample size is less than 30. For small sample sizes, we must ensure that the data is normally distributed to proceed. Since we lack information confirming normality, we cannot construct a confidence interval in this case.

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

In the final scenario, we have 25 randomly selected students spending 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 (\(s\)), we must use the t distribution, applying the critical value \(t_{\alpha/2}\) to construct a 98% confidence interval.

In summary, the decision to use the standard normal or t distribution hinges on 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 for population means.

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 working 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. The first step in constructing the confidence interval is to ensure 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 \(n = 36\), both conditions are satisfied.

Next, we determine the critical t-value. For a 90% confidence level, the significance level (\(\alpha\)) is \(1 - 0.9 = 0.1\), 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 the critical t-value (\(t_{\alpha/2}\)) is approximately \(1.690\).

Now, we calculate the margin of error (\(E\)). The formula for the margin of error when using the sample standard deviation is:

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

Substituting the values, we have:

\[ E = 1.690 \times \frac{21}{\sqrt{36}} \]

Calculating this gives a margin of error of approximately \(5.915\) minutes.

To find the confidence interval, we add and subtract the margin of error from the sample mean. Thus, the lower bound is:

\[ 60 - 5.915 = 54.085 \text{ minutes} \]

And the upper bound is:

\[ 60 + 5.915 = 65.915 \text{ minutes} \]

Therefore, the 90% confidence interval for the true population mean travel time is between \(54.085\) minutes and \(65.915\) minutes. This means we can be 90% confident that the average time to travel to work falls within this range.