Statistics Quiz with Solutions (5 Questions)

The parameter \( \text{df}_2 \) in the F-distribution is often referred to as the **degrees of freedom for the denominator**.

In the context of an F-distribution, there are two sets of degrees of freedom:
- \( \text{df}_1 \): the degrees of freedom for the numerator (associated with the variance estimate in the numerator of the F-ratio).
- \( \text{df}_2 \): the degrees of freedom for the denominator (associated with the variance estimate in the denominator of the F-ratio).

The F-distribution is typically used in the context of ANOVA (Analysis of Variance) and regression analysis to compare variances and test hypotheses about the equality of multiple group means.

To find the probability that the battery life for a randomly selected cell phone is below 3 hours, given that the battery life is normally distributed with a mean (\(\mu\)) of 4 hours and a standard deviation (\(\sigma\)) of 0.8 hours, we need to calculate the corresponding z-score and then find the probability associated with that z-score from the standard normal distribution.

The z-score is calculated using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where:
- \( X \) is the value of interest (3 hours),
- \( \mu \) is the mean (4 hours),
- \( \sigma \) is the standard deviation (0.8 hours).

Plugging in the values, we get:
\[
z = \frac{3 - 4}{0.8} = \frac{-1}{0.8} = -1.25
\]

Next, we look up the z-score of -1.25 in the standard normal distribution table or use a calculator to find the probability that a z-score is less than -1.25.

The cumulative probability for \( Z = -1.25 \) is approximately 0.1056.

Therefore, the probability that the battery life for a randomly selected cell phone is below 3 hours is approximately **0.1056** or **10.56%**.

For a sample proportion \( p \) to have an approximately normal distribution, one of the conditions is that the sample size \( n \) is large enough such that both \( np \) and \( n(1 - p) \) are greater than 5. This ensures that the distribution of the sample proportion can be approximated by a normal distribution.

The condition above involves the range of plausible values for the sample proportion \( p \). Specifically, we want to ensure that the interval \([p - 3\sigma_p, p + 3\sigma_p]\) lies wholly within the interval \([0, 1]\). Here, \( \sigma_p \) is the standard error of the sample proportion, given by:

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

This condition ensures that the sample proportion remains within the valid probability range of 0 to 1.

The interval \([p - 3\sigma_p, p + 3\sigma_p]\) should lie wholly within the interval **[0, 1]**.

In other words, for the sample proportion \( p \) to be approximately normal, the interval:

\[
[p - 3\sigma_p, p + 3\sigma_p]
\]

must be entirely contained within the interval:

\[
[0, 1]
\]

To determine the probability that a randomly selected student spends more than $300 for textbooks each semester, we need to use the properties of the normal distribution.

Given:
- Mean (\(\mu\)): $250
- Standard deviation (\(\sigma\)): $50
- Amount (\(X\)): $300

We first need to convert the given amount to a z-score using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]

Plugging in the values:
\[
z = \frac{300 - 250}{50} = \frac{50}{50} = 1
\]

Next, we look up the z-score of 1 in the standard normal distribution table to find the probability that \(Z \leq 1\).

The cumulative probability for \(Z = 1\) is approximately 0.8413. This value represents the probability that a randomly selected student spends less than $300 for textbooks.

To find the probability that a student spends more than $300, we need to subtract this cumulative probability from 1:
\[
P(X > 300) = 1 - P(X \leq 300) = 1 - 0.8413 = 0.1587
\]

Therefore, the probability that a randomly selected student spends more than $300 for textbooks each semester is approximately **0.1587** or **15.87%**.

To estimate the mean time that employees take to complete an assigned project with 99.8% confidence to within 1 day, we need to determine the minimum sample size required.

Given:
- Confidence level: 99.8%
- Margin of error (\(E\)): 1 day
- Standard deviation (\(\sigma\)): 3 days

We use the formula for the sample size \(n\) for estimating a mean:
\[
n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2
\]

First, we need to determine the critical value \(Z_{\alpha/2}\) for a 99.8% confidence level. The confidence level of 99.8% means that \(\alpha = 1 - 0.998 = 0.002\). Thus, \(\alpha/2 = 0.001\).

We find \(Z_{\alpha/2}\) using the standard normal distribution table or a calculator:
\[
Z_{\alpha/2} \approx 3.090
\]

Next, we plug in the values into the formula:
\[
n = \left( \frac{3.090 \cdot 3}{1} \right)^2
\]

Simplify the expression:
\[
n = \left( \frac{9.27}{1} \right)^2 = 9.27^2 = 85.9329
\]

Since the sample size must be a whole number, we round up to the nearest whole number:
\[
n \approx 86
\]

Therefore, the minimum sample size required to estimate the mean time that employees take to complete an assigned project with 99.8% confidence to within 1 day is **86**.