Question

A random sample of 429 college students was interviewed about a number of matters. a. They reported that they had spent an average of $\mathrm{\$}345.23$ on textbooks during the previous semester. If the sample standard deviation for these data is $\mathrm{\$}15.78$, construct an estimate of the population mean at the $99\mathrm{\%}$ level. b. They also reported that they had visited the health clinic an average of $1.5$ times a semester. If the sample standard deviation is $0.3$, construct an estimate of the population mean at the $99\mathrm{\%}$ level. c. On the average, the sample had missed $2.8$ days of classes per semester because of illness. If the sample standard deviation is $1.0$, construct an estimate of the population mean at the $99\mathrm{\%}$ level. d. On the average, the sample had missed $3.5$ days of classes per semester for reasons other than illness. If the sample standard deviation is $1.5,\mathrm{con}-$ struct an estimate of the population mean at the $99\mathrm{\%}$ level.

Short answer

a. [343.27, 347.19], b. [1.463, 1.537], c. [2.676, 2.924], d. [3.314, 3.686]

Step-by-step solution

Title - Understand Confidence Interval Formula

For each part, the confidence interval for the population mean is given by $$\overline{x}\pm z\ast \frac{s}{\sqrt{n}}$$. Here, $\overline{x}$ is the sample mean, $z$ is the z-value corresponding to the confidence level, $s$ is the sample standard deviation, and $n$ is the sample size.

Title - Find the z-value for 99% Confidence Interval

From the standard normal distribution table, the z-value for a 99% confidence level is approximately 2.576.

Title - Calculate the Confidence Interval for Part a

Given: $\overline{x}=345.23$, $s=15.78$, and $n=429$. The margin of error (E) is $$E=z\ast \frac{s}{\sqrt{n}}=2.576\frac{15.78}{\sqrt{429}}\approx 1.96$$. Thus, the confidence interval is $$345.23\pm 1.96=[343.27,347.19]$$.

Title - Calculate the Confidence Interval for Part b

Given: $\overline{x}=1.5$, $s=0.3$, and $n=429$. The margin of error (E) is $$E=2.576\frac{0.3}{\sqrt{429}}\approx 0.037$$. Thus, the confidence interval is $$1.5\pm 0.037=[1.463,1.537]$$.

Title - Calculate the Confidence Interval for Part c

Given: $\overline{x}=2.8$, $s=1.0$, and $n=429$. The margin of error (E) is $$E=2.576\frac{1.0}{\sqrt{429}}\approx 0.124$$. Thus, the confidence interval is $$2.8\pm 0.124=[2.676,2.924]$$.

Title - Calculate the Confidence Interval for Part d

Given: $\overline{x}=3.5$, $s=1.5$, and $n=429$. The margin of error (E) is $$E=2.576\frac{1.5}{\sqrt{429}}\approx 0.186$$. Thus, the confidence interval is $$3.5\pm 0.186=[3.314,3.686]$$.

Key concepts

Sample Mean

The sample mean, often represented as $\overline{x}$, is a crucial concept in statistics. It refers to the average value obtained from a sample, a subset of a population. For example, if we have data on how much 429 college students spent on textbooks, the sample mean would be the average money spent. It provides a good estimate of the population mean, though some variability should be expected since it's based on a sample. By summing all individual values in the sample and then dividing by the number of observations (n), we get the sample mean. This value is foundational in constructing confidence intervals, as shown in the exercise.

Z-Value

The z-value, or z-score, is a measure used in statistics to indicate how many standard deviations a particular data point is from the mean of the distribution. In the context of confidence intervals, the z-value helps determine the margin of error. When constructing a confidence interval, we multiply the z-value by the standard error to find this margin. For example, in a 99% confidence interval, a z-value of approximately 2.576 is used. This value comes from the standard normal distribution table, signifying that we are confident that 99% of the data points lie within 2.576 standard deviations from the mean.

Standard Deviation

Standard deviation, denoted as $s$, measures the spread of a set of numbers around the mean. It's crucial for understanding the variability within a sample. In the provided exercise, standard deviations for various questions were given (e.g., $15.78 for textbook spending). It helps in calculating the confidence interval by associating it with the sample mean. The lower the standard deviation, the closer the data points are to the mean, and vice versa. To compute it, one can use the formula: $$s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n-1}}$$ where $x_i$ represents each value in the sample, $\overline{x}$ is the sample mean, and $n$ is the number of observations.

Population Mean

The population mean, denoted as $\mu$, is the average of all the values in a population. It's often estimated using the sample mean ($\overline{x}$). In many scenarios, like the exercise provided, the exact population mean is unknown. That's why we use samples to make estimates. The closer the sample mean to the population mean, the better our estimates become. When we construct confidence intervals, we aim to say that the population mean falls within a specific range with a certain level of confidence. For instance, if a confidence interval for textbook spending is [343.27, 347.19], we are 99% confident that the true population mean lies within this range.

Margin of Error

The margin of error (E) quantifies the range within which the true population parameter will lie, considering a specific level of confidence. It's derived by multiplying the z-value by the standard error, which is the standard deviation divided by the square root of the sample size ($\text{(}s/\sqrt{n}$\)). For example, in the provided solutions, the margin of error for the textbook spending was calculated as: $$E=2.576\frac{15.78}{\sqrt{429}}\approx 1.96$$. This value, when added to and subtracted from the sample mean, forms the bounds of the confidence interval. Thus, the margin of error is essential for understanding the precision of our estimates.