Statistics Quiz with Solutions (5 Questions)

To determine the \(F_{0.025}\) test statistic value for degrees of freedom \(df_1 = 5\) and \(df_2 = 7\), we are referring to the F-distribution's critical value at the 0.025 significance level, with \(df_1\) and \(df_2\) representing the degrees of freedom for the numerator and denominator, respectively. This critical value is used in the context of hypothesis testing to decide whether to reject the null hypothesis, specifically in ANOVA or certain regression analyses where the F-test is applicable.

The \(F_{0.025}\) value is found in F-distribution tables or can be computed using statistical software. This value serves as a threshold: if the calculated F-statistic from your data exceeds this value, it indicates that there is a statistically significant difference between the groups being compared at the 0.025 significance level.

The \(F_{0.025}\) test statistic value for degrees of freedom \(df_1 = 5\) and \(df_2 = 7\) is approximately \(5.285\). This means that, at the 0.025 significance level, if your calculated F-statistic exceeds \(5.285\), there is a statistically significant difference between the groups you are comparing, leading to the rejection of the null hypothesis in the context of an ANOVA test or a similar statistical analysis.

The description provided, involving the summary of sample size, sample sizes in select subgroups, demographics like age, and the proportion of subjects of each breed, refers to activities in the realm of **Descriptive Statistics**. Descriptive statistics is the branch of statistics that deals with the organization, summarization, and visualization of data to describe the basic features of the data in a study. It provides simple summaries about the sample and the measures. Through descriptive statistics, researchers can present quantitative descriptions in a manageable form, involving measures of central tendency (like mean, median, mode), measures of variability (like range, variance, standard deviation), and frequency distributions, including proportions and percentages. This branch is crucial for the initial analysis of data, setting the stage for further statistical analysis or inferential statistics, where conclusions about the population are drawn from the sample data.

Given that \(A\) and \(B\) are independent events, the probability of both \(A\) and \(B\) occurring, denoted as \(P(A \cap B)\), can be calculated using the formula for the probability of the intersection of two independent events:

\[P(A \cap B) = P(A) \times P(B)\]

However, we are initially provided with \(P(A^c)\), the probability of the complement of \(A\), and \(P(B)\). Since \(P(A) + P(A^c) = 1\), we can find \(P(A)\) as:

\[P(A) = 1 - P(A^c)\]

Substituting \(P(A^c) = \frac{1}{5}\) into the equation gives us \(P(A)\):

\[P(A) = 1 - \frac{1}{5} = \frac{4}{5}\]

Now, using \(P(A) = \frac{4}{5}\) and the given \(P(B) = \frac{1}{8}\), we calculate \(P(A \cap B)\) as:

\[P(A \cap B) = \frac{4}{5} \times \frac{1}{8} = \frac{4}{40} = \frac{1}{10}\]

Therefore, the probability of both \(A\) and \(B\) occurring, \(P(A \cap B)\), is \(\frac{1}{10}\).

To compute the sample median for the given data set \(\{-5, 4, 8, 5, 2, -1, 3, -6\}\), we first need to sort the data set in ascending order, which results in \(\{-6, -5, -1, 2, 3, 4, 5, 8\}\). The sorted data set has an even number of observations (\(N = 8\)).

For data sets with an even number of observations, the median is calculated as the average of the two middle numbers. In this case, the middle numbers are the fourth and fifth observations in the sorted list, which are \(2\) and \(3\), respectively.

Thus, the median is calculated as follows:

\[
\text{Median} = \frac{2 + 3}{2} = 2.5
\]

Therefore, the sample median of the given data set is \(2.5\).

To construct a 99% confidence interval for the proportion of all students in the university who like to play football, given a random sample of 300 students where 75 declared their interest in playing football, we use the formula:

\[CI = \hat{p} \pm Z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]

where:
- \(\hat{p}\) is the sample proportion, calculated as \(\hat{p} = \frac{75}{300}\),
- \(Z_{\frac{\alpha}{2}}\) is the Z-score for the 99% confidence level, which is approximately \(2.576\),
- \(n\) is the sample size, which is \(300\).

Substituting the given values into the formula, we get:

\[\hat{p} = \frac{75}{300} = 0.25\]

The confidence interval is calculated as:

\[CI = 0.25 \pm 2.576 \cdot \sqrt{\frac{0.25 \times (1 - 0.25)}{300}}\]

After the calculations, the confidence interval is found to be approximately:

\[CI \approx (0.186, 0.314)\]

This means we are 99% confident that the true proportion of all students in the university who like to play football is between 18.6% and 31.4%.