Statistics Quiz with Solutions (10 Questions)

A census is preferred over a sample when it's feasible to measure or observe every member of the population of interest, especially when the population is small or when accurate, comprehensive data is critical.

For a value \(x\) in a data set, if \(P\) percent of the values are less than or equal to \(x\), then \(x\) is called the **Pth percentile**.

The term "population" refers to the complete set of all items that share at least one property of interest. In contrast, a "sample" is a subset of the population, usually selected for the purpose of observation and analysis. Based on these definitions:

**Heights of all students in a college** represents a population because it includes the entire group of interest (all students in the college) without exception.

A random experiment is defined as an experiment or a process for which the outcome cannot be predicted with certainty in advance. An experiment conducted to verify Archimedes' principle is not a random experiment. Archimedes' principle is a law of physics that describes the buoyant force on an object submerged in a fluid. Experiments to verify it are designed to observe this principle in action, rather than to obtain unpredictable outcomes. The results of such experiments, when conducted correctly, are expected and based on established physical laws, not on randomness.

To find \(P(\hat{p} \leq 0.6)\), where \(\hat{p}\) is the sample proportion, we're interested in calculating the probability that the sample proportion is less than or equal to 0.6 given a population proportion \(p = 0.56\) and a sample size of \(n = 120\). To do this, we'll use the Z-score formula for a sample proportion, which is:

\[
Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}
\]

In this case, \(\hat{p} = 0.6\), \(p = 0.56\), and \(n = 120\). Let's calculate the Z-score and then use it to find the probability \(P(\hat{p} \leq 0.6)\).

The Z-score for a sample proportion of \(\hat{p} = 0.6\) is approximately \(0.883\), and the probability \(P(\hat{p} \leq 0.6)\) is approximately \(81.13\%\). This means there's an 81.13% chance that the sample proportion will be 0.6 or less, given a population proportion of 0.56 with a sample size of 120.

To construct an 80% confidence interval for the mean time spent in the beauty salon by all the women in the city, we use the formula for a confidence interval for the mean, which is:

\[
\bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

where:
- \(\bar{x}\) is the sample mean,
- \(Z_{\alpha/2}\) is the Z-score corresponding to the desired confidence level (in this case, 80%),
- \(\sigma\) is the standard deviation of the sample, and
- \(n\) is the sample size.

Given:
- \(\bar{x} = 44\) minutes,
- \(\sigma = 16\) minutes, and
- \(n = 200\).

First, we need to find \(Z_{\alpha/2}\) for an 80% confidence level, which means \(\alpha = 1 - 0.80 = 0.20\), so \(\alpha/2 = 0.10\). We'll then calculate the confidence interval.

For an 80% confidence interval, the Z-score (\(Z_{\alpha/2}\)) is approximately \(1.282\). The margin of error is about \(1.45\) minutes.

Therefore, the 80% confidence interval for the mean time spent in the salon by all the women in the city is approximately \(42.55\) minutes to \(45.45\) minutes. This interval suggests that we can be 80% confident that the true mean time spent in the beauty salon by all women in the city lies within this range.

When constructing confidence intervals, especially for proportions, the length of the interval is influenced by the confidence level and the variability within the sample, but not directly by degrees of freedom, which are more relevant to t-distributions used for means with smaller sample sizes.
A higher confidence level means you want to be more certain about capturing the true parameter, which requires widening the interval. As confidence levels increase (e.g., from 90% to 95% to 99%), the corresponding Z-scores also increase, leading to a longer confidence interval.

To calculate the percentage of students who will be promoted to the next grade, we utilize the properties of the normal distribution. The normal distribution is defined by two parameters: the mean \(\mu\) and the standard deviation \(\sigma\). For this scenario, we have \(\mu = 60\) and \(\sigma = 5.5\), and the promotion cutoff score is \(55\).

The percentage of students scoring above the cutoff can be found by calculating the area under the normal distribution curve to the right of the cutoff score. This area represents the probability that a randomly selected student's score exceeds the cutoff. The cumulative distribution function (CDF) gives us the probability that a variable takes a value less than or equal to a certain value in a distribution. Therefore, to find the probability of scoring above the cutoff, we subtract the CDF at the cutoff score from 1:

\[P(X > 55) = 1 - P(X \leq 55) = 1 - F(55)\]

where \(F(55)\) is the CDF for a normal distribution with mean \(\mu = 60\) and standard deviation \(\sigma = 5.5\) evaluated at \(55\).

In mathematical terms, this calculation is:

\[P(X > 55) = 1 - \Phi\left(\frac{55 - \mu}{\sigma}\right)\]

where \(\Phi\) denotes the standard normal CDF. After computing this expression, we found that approximately \(81.83\%\) of the students will be promoted to the next grade, as this percentage of students scores above \(55\).